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.