Subring of $\mathcal O(\mathbb C)$

$\def\fA{\mathfrak{A}}\def\cO{\mathcal{O}}$In this answer we will show that $\fA$ does not contain the identity function $z \mapsto z$. As already sketched in my comments above, this will imply that $\fA$ does not contain any function which has a finite, but non-zero, number of zeroes.

The key is an 1897 theorem of Borel due to uniqueness of representations of the above form. A good modern reference is Chapter VII in Lang's Introduction to Complex Hyperbolic Spaces.

Define an equivalence relation on $\cO$ by $f \sim g$ iff $f-g$ is the constant function. Suppose that $\sum_{i=1}^n \exp(f_i) = 0$ for $f_1$, $f_2$, ..., $f_n \in \cO$. Partition $\{ f_1, f_2, \ldots, f_n \}$ into equivalence classes. Then Borel's result (Theorem 1.1 in Lang) says that, for each equivalence class $S$, we have $\sum_{f_i \in S} \exp(f_i)=0$. Since each equivalence class is of the form $(h+c_1, h+c_2, \dots, h+c_k)$ for some $h \in \cO$ and some constants $c_1$, ..., $c_k$, this is saying that essentially the only solution to $\sum \exp(f_i)=0$ is to look at $\sum \exp(h+c_i)$ where $\sum \exp(c_i)=0$. In particular, there are no singleton equivalence classes.

Let us reframe Borel's theorem. Let's say that a sum $\sum \exp(f_i)$ is in reduced form if no two $f_i$ are equivalent. We can always put any sum into reduced form by replacing $\exp(h+c_1)+\exp(h+c_2)$ by $\exp(h+\log(e^{c_1}+e^{c_2}))$ if $e^{c_1}+e^{c_2} \neq 0$, or by the empty sum if $e^{c_1}+e^{c_2}=0$.

Theorem If $\sum_{a=1}^m \exp(f_a) = \sum_{b=1}^n \exp(g_b)$ with both sums in reduced form, then $m=n$ and the $f_a$ and $g_b$ are the same up to permutation and adding integer multiple of $2 \pi i$.

Proof: Partition $\{ f_1, \ldots, f_m, g_1+i \pi, \ldots, g_n+i \pi \}$ into equivalence classes. By the hypothesis that the sums are in reduced form, no equivalence class contains two $f_a$'s or two $g_b$'s.

Note that $\sum \exp(f_a) + \sum \exp(g_b + \pi i)=0$. By Borel's theorem, no equivalence class is a singleton. So all equivalence classes are of the form $\{ f_a, g_b \}$. In particular, $m=n$ and there is some permutation $\sigma$ and some constants $c_a$ such that $g_{\sigma(a)} +i \pi = f_a + c_a$. Now, Borel's theorem further tells us that $\exp(i \pi) + \exp(c_a)=0$. So $c_a$ is an integer multiple of $2 \pi i$. $\square$

We will now use this theorem to show that $z \not \in \fA$. Suppose for the sake of contradiction that $z = \sum_{a=1}^n \exp(f_a(z))$, and assume the sum is in reduced form.

Then, for any nonzero constant $h$, we have $$\sum_{a=1}^n \exp(f_a(z))+\exp(\log h) = z+h = \sum_{a=1}^n \exp(f_a(z+h)).$$ The right hand side is in reduced form. If none of the $f_a$ are constants, then the left hand side is in reduced form, and we have a reduced sum of $n+1$ terms equaling a reduced sum of $n$ terms, a contradiction. If one of the $f_a$ is a constant, cancel that term from both sides and once again reach a contradiction.

Remark As my comments show above, this rules out any function with finitely many zeroes. I think it should also rule out things like $f(z) = z(2^z+3^z)$, by considering the relation $$f(z+2) - 5 f(z+1) + 6 f(z) = 3 \cdot 3^{z} - 2 \cdot 2^z.$$ However, I do not have a good criterion for when a function is in $\fA$.

Remark Choose one function from each equivalence class. Then Borel's theorem shows that $\fA$ is the free $\mathbb{C}$ vectors space on the exponentials of these representatives. As a ring, $\fA$ is the group algebra of the abelian group $\cO^+/\mathbb{C} \cdot 1$, where $\cO^+$ means the additive group of the ring $\cO$.

Remark Note that $\fA$ is not closed under natural limiting operations. For example, for any $h \neq 0$, the function $z \mapsto \frac{e^{hz}-1}{h}$ is in $\fA$. But $\lim_{h \to 0} \frac{e^{hz}-1}{h} = z$ is not, even though the limit is uniform on compact subsets.