I posted a bunch of comments here, but I thought it would be better to write an answer, instead.

First, I've written up a lot about these functions, and functions with the same condition on sets $S\subseteq \mathbb Z$.

If you define $$q_k(x) = \frac{\operatorname{lcm}(1,2,\dots,k)}{k!}\left(x+ \left\lfloor {\frac{k-1}{2}} \right\rfloor\right)_k$$ then these are rational polynomials in $R$ and they have the property that, for all $n\in\mathbb Z$, $q_k(n)\neq 0$ for only finitely many $k$. So given any infinite sequence $(a_k)_{k\in\mathbb N}$ of integers, we get a unique $f\in R$ defined by:

$$f(n)=\sum_{k} a_kq_k(n)$$

A little harder to prove, but not really hard, is that every $f\in R$ is expressed in this form. Essentially, you find $a_k$ by induction.

This gives an explicit abelian group isomorphism between $R$ and $\mathbb Z^{\mathbb N}\cong\mathbb Z^{\mathbb Z}$.

The proof is essentially using Chinese remainder there inductively to determine $a_k$, to match the values of $f(0),f(1),f(-1),f(2),f(-2),\dots.$ This is why you get the term $\operatorname{lcm}(1,2,\dots,k)$.

This also helps show that $R\cap \mathbb Q[x]$ has the $q_k$ as $\mathbb Z$-generators. Given a polynomial $q(x)\in Q[x]\cap R$ is of degree $d$, if $q(n)=\sum_{k\in N} a_k q_k(n)$ then $p(n)=\sum_{k=0}^{d} a_kq_k(n)$ is another polynomial of degree $d$ which agrees at $d+1$ values, so must agree everywhere.

Side note: $R$ is actually an integral domain, not just a ring. If $f\in R$, then $f$ can be zero only for finitely many values, so $fg=0$ implies $f=0$ or $g=0$.

Using $R'=\left\{f:\mathbb N\to \mathbb Z\mid \forall n,m: n-m\mid f(n)-f(m)\right\}$, there is a natural ring homomorphism $R\to R'$ be sending $f\in R$ to $f_{\mid\mathbb N}\in R'$. This is $1-1$ since if $f_{\mid\mathbb N} =0$, then $f$ has infinitely many zeros. The map is not onto - it is possible to define $f\in R'$ such that you can't even extend $f$ to $-1$ and keep the divisibliity property.