Is the ring and semi-ring definition of algebra and set linked?

Using your definition of "ring of sets in measure theory", that it is a (nonempty) collection of sets $R$ such that (1) it is closed under union ($\forall A,B\in R$ we have $A\cup B\in R$) and (2) it it is closed under set-theoretic difference ($\forall A,B\in R$ we have $A-B\in R$), we can in fact show $R$ is a commutative ring in the algebra sense, however with respect to the operations symmetric difference and intersection. So first we show that in fact, any ring of sets in measure theory is closed over symmetric difference and intersection:

Lemma. Let $R$ be a ring of set, then for all $A,B\in R$, we have $A\Delta B\in R$ and $A\cap B\in R$. (Here $A\Delta B:=(A-B)\cup(B-A)$ is the operation symmetric difference.)

Pf. The closure of symmetric difference is clear from its definition. And note well that $A\cap B= (A\cup B)-(A\Delta B)$. $\Box$

Now we make the observation: If $R$ is a ring of sets, then $\varnothing\in R$, since for any $A\in R$, we have $A\Delta A=\varnothing \in R$. Further note that for any $A\in R$, we have $\varnothing\Delta A=A$. Hence if we identify the operation $\Delta$ as "addition" and $\varnothing$ as the "additive identity", and every $A\in R$ is its own "additive inverse", we have

Claim. If $R$ is a ring of sets, then $(R,\Delta)$ is an abelian gorup. $\Box$ (Associativity given set-theoretically.)

Now, identify the operation $\cap$ as "multiplication", then with the following

Fact. Intersection distributes over symmetric difference. $\Box$

we have finally:

Proposition. If $R$ is a ring of set (measure theory sense), then it is also closed under symmetric difference $\Delta$ and intersection $\cap$. And that $(R,\Delta,\cap)$ forms a commutative ring (algebra sense). $\Box$

This rather long answer is meant to expand on @bonsoon's good answer by presenting the deeper reason why set-theoretic symmetric difference and intersection are precisely the operations we need to prove that measure-theoretic rings of subsets are also rings in the algebraic sense.

(From now on, I use the denomination "algebraic ring" to refer to the ring structure as defined in abstract algebra, and "measure-theoretic ring" to refer to the measure-theoretic definition of "ring of subsets of a set".)

Theorem. Let $A \neq \varnothing$ be a set. A family $\mathcal R$ of subsets of $A$ may form an algebraic ring of subsets of $A$ if and only if it is a measure-theoretic ring of subsets of $A$, i.e. it is such that

• $\forall X,Y \in \mathcal R \qquad X \bigtriangleup Y \in \mathcal R$
• $\forall X,Y \in \mathcal R \qquad X \cap Y \in \mathcal R$

Remark. I still say "may form" because, in principle, we do not know the set-theoretic operations for which the family $\mathcal R$ may be considered an algebraic ring. The appropriate operations will be clear from the following construction.

Preamble. Consider the set $$\mathbb Z_2 = \{0,1\} $$ If $\mathbb Z_2$ is equipped with the binary operations \begin{equation} \begin{split} + : \mathbb Z_2 \times \mathbb Z_2 \to \mathbb Z_2 \qquad &(x,y) \mapsto x+y = (x\ \hat +\ y) \mod 2 \\ \cdot : \mathbb Z_2 \times \mathbb Z_2 \to \mathbb Z_2 \qquad &(x,y) \mapsto x\cdot y = (x\ \hat \cdot\ y) \mod 2 \end{split} \end{equation} where $\hat +$ and $\hat \cdot$ denote the usual addition and multiplication in the integers, then $(\mathbb Z_2,+,\cdot) $ consitutes a (commutative) algebraic ring.

Consider then the set of all maps from the non-empty set $A$ to $\mathbb Z_2$, $$ \mathbb Z_2^A = \{f : A \to \mathbb Z_2\} $$ The generic map $f \in \mathbb Z_2^A$ will send some elements of $A$ to $1 \in \mathbb Z_2$, and the rest to $0 \in \mathbb Z_2$. Call $X\subseteq A$ the set of all values that are mapped to $1$ and denote the map with $\mathbf 1_X$. Consider a similar map $\mathbf 1_Y$ defined analogously from a second subset $Y \subseteq A$. (These maps are called characteristic functions of sets $X$ and $Y$.) We want to induce the above-defined binary operations over $\mathbb Z_2$ upon the set $\mathbb Z_2^A$ in the following way: for all $t \in A$, set \begin{equation} \begin{split} \oplus : \mathbb Z_2^A \times \mathbb Z_2^A \to \mathbb Z_2^A \qquad &(\mathbf 1_X \oplus \mathbf 1_Y)(t) = \mathbf 1_X(t) +\mathbf 1_Y(t) \quad \forall t \in A \\ \odot : \mathbb Z_2^A \times \mathbb Z_2^A \to \mathbb Z_2^A \qquad &(\mathbf 1_X \odot \mathbf 1_Y)(t) = \mathbf 1_X(t)\cdot\mathbf 1_Y(t) \quad \forall t \in A \end{split} \end{equation} This way, $(\mathbb Z_2^A,\oplus,\odot)$ too becomes a (commutative) algebraic ring, where the identity element for $\oplus$ is $\mathbf 1_\varnothing$ (the identically zero map on $A$), and the identity element for $\odot$ is $\mathbf 1_A$.

Relations with set-theoretic operations. Consider two arbitrary subsets $X,Y$ of $A$. We see trivially that $$ (\mathbf 1_X \oplus \mathbf 1_Y)(t) = \begin{cases} 0 + 0 = 0 & \forall t \in A \setminus (X \cup Y)\\ 1 + 0 = 1 & \forall t \in X \setminus Y \\ 0 + 1 = 1 & \forall t \in Y \setminus X \\ 1 + 1 = 0 & \forall t \in X \cup Y \\ \end{cases} $$ and that $$ (\mathbf 1_X \oplus \mathbf 1_Y)(t) = \begin{cases} 0 \cdot 0 = 0 & \forall t \in A \setminus (X \cup Y)\\ 1 \cdot 0 = 0 & \forall t \in X \setminus Y \\ 0 \cdot 1 = 0 & \forall t \in Y \setminus X \\ 1 \cdot 1 = 1 & \forall t \in X \cup Y \\ \end{cases} $$ Therefore, simply by applying the rules for addition and multiplication in $\mathbb Z_2$, one can check that $$ (\mathbf 1_X \oplus \mathbf 1_Y)(t) = \mathbf 1_X(t) +\mathbf 1_Y(t) = \mathbf 1_{X \bigtriangleup Y}(t), \qquad \forall t \in A $$ and that $$ (\mathbf 1_X \odot \mathbf 1_Y)(t) = \mathbf 1_X(t)\cdot\mathbf 1_Y(t) = \mathbf 1_{X\cap Y}(t), \qquad \forall t \in A $$

The natural isomorphism. We can now identify $\mathbb Z_2^A$ with $\mathcal P(A)$ through the natural isomorphism $\mathfrak I : \mathbf 1_X \mapsto X$, concluding that $(\mathcal P(A),\bigtriangleup,\cap)$ must be an algebraic ring of subsets of $A$, where the additive operation coincides with set-theoretic symmetric difference and has identity $\varnothing$, and the multiplicative operation coincides with set-theoretic intersection and has identity $A$.

Subrings. The existence of the ring isomorphism $\mathfrak I$ allows us to say that each subring of $(\mathbb Z_2^A, \oplus, \odot)$ is isomorphic to a subring of $(\mathcal P(A), \bigtriangleup, \cap)$, which is then an algebraic ring of subsets of $A$ with respect to $\bigtriangleup$ and $\cap$. Thus we recognize these two operations to be the ones we need to complete our proof.

Characterization of algebraic rings of subsets. We may describe the nature of algebraic rings of subsets of $A$ by studying that of subrings of $\mathbb Z_2^A$. Let $R \subseteq \mathbb Z_2^A$ be non-empty and $(R,\oplus,\odot)$ be a subring of $(\mathbb Z_2^A,\oplus,\odot)$. By the subring criterion (proof here), this is equivalent to saying that $R$ is closed under $\odot$, it contains $\mathbf 1_\varnothing$ and the $\oplus$-inverses of each of its elements. The last requirement means that for all $\mathbf 1_X \in R$, there exists $\mathbf 1_Y \in R$ such that $\mathbf 1_X \oplus \mathbf 1_Y = \mathbf 1_\varnothing$. If $\mathbf 1_X = \mathbf 1_\varnothing$ this surely holds; if $\mathbf 1_X \neq \mathbf 1_\varnothing$, we have $$ (\mathbf 1_X \oplus \mathbf 1_Y)(t) = \mathbf 1_X(t) + \mathbf 1_Y(t) = \mathbf 1_\varnothing(t) = 0 \quad \forall t \in A $$ This tells us $\mathbf 1_Y(t)$ should be the additive inverse of $\mathbf 1_X(t)$ in $\mathbb Z_2$ for all $t$ in $A$. By the fact that the additive inverse of all (both) $x \in \mathbb Z_2$ is $x$ itself, it follows that $\mathbf 1_Y(t) = \mathbf 1_X(t)$ for all $t$ in $A$, and so $\mathbf 1_Y = \mathbf 1_X$. This means that any subset of $\mathbb Z_2^A$ may constitute a subring of $(\mathbb Z_2^A, \oplus,\odot)$ as long as it is closed under $\odot$ and it contains the element $\mathbf 1_\varnothing$! When we apply our ring isomorphism $\mathfrak I$ to $R$, we find a family $\mathcal R$ of subsets of $A$ that is closed under $\cap$ and contains $\varnothing$: these conditions are sufficient for it to be an algebraic ring of subsets of $A$.

We may now proceed to the actual proof.

Proof. Let us prove the leftward implication. Let $\mathcal R \subseteq \mathcal P(A)$ be a measure-theoretic ring of subsets of $A$. $\mathcal R$ forms an algebraic ring for the operations $\bigtriangleup$ and $\cap$ if and only if it satisfies the two conditions stated above, viz. it is closed under $\cap$, and it contains $\varnothing$. This is in fact the case: by hypothesis, we see that $\mathcal R$ is closed under intersection; moreover, for any $X \in \mathcal R$, closure under $\bigtriangleup$ provides $X \bigtriangleup X \in \mathcal R$, that is, $\varnothing \in \mathcal R$. Hence we have our thesis.

The rightward implication is trivial: if $(\mathcal R,\bigtriangleup,\cap)$ is an algebraic ring of subsets of $A$, then $\mathcal R$ is closed under both set-theoretic symmetric difference and intersection; therefore it is a measure-theoretic ring of subsets of $A$. $\square$

Remark. Notice that if $\mathcal R$ is a measure-theoretic ring of subsets of a non-empty set $A$, and $A \in \mathcal R$, then $\mathcal R$ is called an algebra of sets. If we look at $\mathcal R$ as an algebraic ring, precisely as a subring of $\mathcal P(A)$, this corresponds to strengthening the condition so that the multiplicative identity of $\mathcal R$ corresponds to that of $\mathcal P(A)$. Some authors actually do require that the multiplicative identity of a subring of a ring be the same as that of the ring in the very definition of subring. If one adopts such a definition of subring (since we have not required here that $A\in \mathcal R$ in general) then it is measure-theoretic algebras, not rings, that have an algebraic ring structure.

Example. Define an $n$-rectangle in $\mathbb R^n$ to be the Cartesian product of $n$ intervals, and call it unbounded if at least one of those intervals is unbounded, bounded otherwise. Define an $n$-multirectangle to be the union of a finite number of $n$-rectangles, and call it unbounded if at least one of the $n$-rectangles is unbounded, bounded otherwise (so all $n$-rectangles are also $n$-multirectangles). The set of all bounded $n$-multirectangles in $\mathbb R^n$ constitutes an algebraic ring of subsets of $\mathbb R^n$, but does not contain $\mathbb R^n$ itself. Instead, if we allow the $n$-multirectangles to be unbounded, we retrieve $\mathbb R^n$ as $\mathbb R^n = (-\infty,+\infty)^n$, and our ring becomes an algebra.