This is true for any abelian groups (finitely generated and free abelian is not needed); see Corollary 3 of [this paper][1].  Basically, if $\mathbb ZG_1\cong \mathbb ZG_2$, then $H_n(G_1,\mathbb Z)\cong H_n(G_2,\mathbb Z)$ for all $n\geq 0$ and since $H_1$ gives the abelianization, this gives the desired result.  In fact, isomorphism of group rings implies isomorphism of abelianization.  For free abelian groups, this has many elementary proofs.  The easiest to me is that Laurent polynomial rings have only trivial units (multiples of group elements) by looking at highest and lowest degree terms in a unit times its inverse and it is easy to prove that if a group ring $\mathbb ZG$ has only trivial units then $G$ is recovered as $(\mathbb ZG)^{\times}/\mathbb Z^\times$.

The key idea for the homology approach  is to show that every isomorphism of groups rings can be replaced by one which preserves augmentation.  It suffices to show that if $\omega\colon \mathbb ZG\to\mathbb Z$ is the augmentation map and $\alpha\colon \mathbb ZG\to \mathbb Z$ is any ring homomorphism then there is an automorphism $\sigma$ of $\mathbb ZG$ with $\alpha\sigma=\omega$.  But $\alpha(g)\in \mathbb Z^\times$ and so $\sigma(g) = \alpha(g)^{-1}g$ is the desired automorphism.


  [1]: https://www.ams.org/journals/tran/1969-136-00/S0002-9947-1969-0233903-9/S0002-9947-1969-0233903-9.pdf