A commutative group structure on $R\times R$ for a ring $R$

With $R=\mathbb{Z}$, you get an isomorphism. Note that $(\mathbb{Z}\times\mathbb{Z},\circ)$ is torsionfree abelian: given $(a,b)\in\mathbb{Z}\times\mathbb{Z}$, if $a\neq 0$, then $n(a,b) = (na,x)$ for some $x$, so $n(a,b)=(0,0)$ requires $n=0$. If $a=0$, then $n(0,b) = (0,nb)$, so again $n(0,b)=0$ requires $n=0$ or $b=0$. (I'm using $m(x,y)$ to mean $(x,y)$ $\circ$-added to itself $m$ times if $m\gt 0$, and the $\circ$-inverse if $m\lt 0$, as usual for an abelian group).

Also, $(\mathbb{Z}\times\mathbb{Z},\circ)$ is $2$-generated: $(1,0)$ and $(0,1)$ certainly generate: to get an arbitrary $(a,b)$ using $(1,0)$ and $(0,1)$, just take $a(1,0)$, which will give an element of the form $(a,x)$ for some $x$, and then take $(a,x)\circ(0,b-x)$.

So $(\mathbb{Z}\times\mathbb{Z},\circ)$ is free abelian and $2$-generated, hence either cyclic or isomorphic to $\mathbb{Z}\times\mathbb{Z}$. Moding out by the infinite cyclic normal subgroup $\{(0,b)\mid b\in\mathbb{Z}\}$ we get a group isomorphic to $\mathbb{Z}$, so the group $(\mathbb{Z}\times\mathbb{Z},\circ)$ cannot be infinite cyclic (because the quotient of an infinite cyclic group by a nontrivial subgroup is finite). Thus, with $R=\mathbb{Z}$ you get an isomorphism.

More generally, if you have an isomorphism for $R$ and one for $S$, then you also get an isomorphism for $R\oplus S$, since the addition will just be "coordinate wise", so you can take the isomorphism $f\colon (R\times R,\circ)\to R\oplus R$ and $g\colon (S\times S,\circ)\to S\oplus S$, and obtain an isomorphism $f\times g\colon (R\oplus S\times R\oplus S,\circ)\to (R\oplus R)\times (S\oplus S)\cong (R\oplus S)\times (R\oplus S)$.

This will give you isomorphisms for all rings of the form $$\mathbb{Z}^t \times\frac{\mathbb{Z}}{p_1^{a_1}\mathbb{Z}}\times\cdots\times\frac{\mathbb{Z}}{p_r^{a_r}\mathbb{Z}}$$ where $p_1,\ldots,p_r$ are primes, different from $3$, $a_i\gt 0$, $t,r\geq 0$, and the ring structure is the obvious one.

Added. (Well that was silly). For rings of characteristic $3$, you get an isomorphism if and only if $a^3=0$ for all $a\in R$.

To see that this is necessary, notice that \begin{align*} 3(a,0) &= (a,0)\circ(a,0)\circ(a,0) = (2a,2a^3)\circ(a,0)\\ &= (3a,2x^3 + 4a^3+2a^3) = (0,8a^3) = (0,-a^3). \end{align*} Thus, $(R\times R,\circ)$ has characteristic $3$ if and only if $a^3=0$ for all $a$.

To see that the condition is sufficient, if $R$ is of characteristic $3$ and $a^3=0$ for all $a\in R$,then $(R\times R,\circ)$ is an abelian $3$-group, hence a vector space over $\mathbb{F}_3$. Since it has the same cardinality as $R\oplus R$, which is also a vector space over $\mathbb{F}_3$, they are isomorphic as $\mathbb{F}_3$-vector spaces, and hence as abelian groups.

Ok, so inspired by Arturo's answer, here is another partial answer that includes the Gaussian integers and the 3-adic integers:

Define τ:R→R:n↦2*(n+1 choose 3) and notice the formal equality

$$2\cdot\binom{n+m+1}{3} = 2\cdot\binom{n+1}{3}+2\cdot\binom{m+1}{3} + (mn^2+m^2n)$$

Then define [ a, b ] = ( a, b + τ(a) ) to be a different coordinate system on (R⊕R,∘). Then [ a1, b1 ]∘[ a2, b2 ] = [ a1+a2, b1+b2 ] is clearly isomorphic to R⊕R.

Suppose R is a domain with field of fractions K. Then τ:R→K:n↦2*(n+1 choose 3) definitely exists, so one just needs some condition for τ(R) ≤ R. I thought this was at least sort of common, but I can't think of any real examples other than Z.

Certainly R=Z[x] doesn't work for τ, though both R⊕R and A will be free abelian of countably infinite rank.

I guess the only improvement is that we don't need every element of R to be divisible by 3, only the elements of the form xxx−x.

I believe the ring R=Z[i] works for this, since Z[i]/(3) ≅ Z/3Z × Z/3Z satisfies xxx-x ≡ 0. This ring is less exciting in some ways though, since the additive group is free abelian.

Rings that are even nicer, where all binomial coefficients exist, are called binomial rings. For example the p-adic integers for any prime p (and where p=3 is the interesting one for us) also work. The additive group of p-adic integers is not free abelian, so this is a real gain.