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



First, I've [written up a lot about these functions][1], 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 expression for an isomorphism between $R$ and $\mathbb Z^{\mathbb N}\cong\mathbb Z^{\mathbb Z}$.

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 $g=0$.

  [1]: http://thomasoandrews.com/math/congruent/