The group algebra $k[G]$ of a finite group $G$ over a field $k$ knows little about $G$ most of the time; if $k$ has characteristic prime to $|G|$ and contains every $|G|^{th}$ root of unity, then $k[G]$ is a direct sum of matrix algebras, one for each irreducible representation of $G$ over $k$, so $k[G]$ only knows the dimensions of the irreducible representations of $G$.
The group ring $\mathbb{Z}[G]$, on the other hand, knows at least as much as $k[G]$ for every $k$ (which one can obtain by tensoring with $k$); this information should let you deduce something about the entries in the character table of $G$, although I'm not sure exactly what. If the characteristic of $k$ divides $|G|$ then $k[G]$ knows something about the modular representation theory of $G$, although again, I'm not sure exactly what. So:
How strong of an invariant is $\mathbb{Z}[G]$? More precisely, what group-theoretic properties of $G$ do I know if I know $\mathbb{Z}[G]$? What are examples of finite groups $G_1, G_2$ which are not isomorphic but which satisfy $\mathbb{Z}[G_1] \cong \mathbb{Z}[G_2]$?
If in addition to $\mathbb{Z}[G]$ we are given the augmentation homomorphism $\mathbb{Z}[G] \to \mathbb{Z}$ then it looks like we can recover the group cohomology and homology of $G$, so it seems plausible that $\mathbb{Z}[G]$ contains quite a lot of information.
At a minimum, setting $k = \mathbb{R}$ I think we can compute the Frobenius-Schur indicator of every complex irreducible representation of $G$.
Edit: It seems the second problem is known as the isomorphism problem for integral group rings and is quite hard.
If you know that $G$ and $H$ are at least abelian, then $\mathbb{Z}[G]\cong\mathbb{Z}[H]$ implies $G\cong H$.
proof: The homology groups of $G$ and $H$ are independent of the choice of resolution up to canonical isomorphism, and the groups are defined by $H_iG=H_i(F_G)$ where $F$ is a projective resolution of $\mathbb{Z}$ over $\mathbb{Z}G$ (similary for $H$). Since $\mathbb{Z}G\cong\mathbb{Z}F$, the $G$-modules $F_i$ can be regarded as $H$-modules via restriction of scalars and hence the projective resolution $F$ for the homology of $G$ can also be used for the homology of $H$. We have $F_G\cong F_H$ since $\mathbb{Z}\otimes_{\mathbb{Z}G}F\cong \mathbb{Z}\otimes_{\mathbb{Z}H}F$ by the obvious map $1\otimes f\mapsto 1\otimes f$ [using the isomorphism $\varphi:\mathbb{Z}G\rightarrow \mathbb{Z}H$ we have $1\otimes f=1\otimes gf\mapsto 1\otimes \varphi(g)f=1\otimes f$], and so $H_i(G)\cong H_i(H)$ for all $i$. In particular, $G/[G,G]\cong H_1(G)\cong H_1(H)\cong H/[H,H]$. Since $G$ and $H$ are abelian groups, $[G,G]=0=[H,H]$ and hence $G\cong H$.
If we knew the augmentation map then we can recover $G_{ab}$ as $I/I^2$, where $I$ is the augmentation ideal (kernel of the augmentation map). But even if we don't know the augmentation map, $\mathbb{Z}[G]\cong\mathbb{Z}[H]$ would imply $I_G\cong I_H$ and hence $G_{ab}\cong H_{ab}$, reproducing my above statement.
And in general there does exist counterexamples, I forgot the sources on them, but this MO question refers to one of them:
https://mathoverflow.net/questions/60609/strong-group-ring-isomorphisms
(I first claimed, without thinking, that $G$ sits as the group of units in $\mathbb{Z}[G]$, which is what Qiaochu refers to in his comment below).