What does the group ring $\mathbb{Z}[G]$ of a finite group know about $G$?

The Finite abelian case-

Let $G_1 $ be a finite abelian group.

If $\Bbb{Z}G_1\cong \Bbb{Z}G_2$, and suppose $\phi$ is the isomorphism, then you can define another isomorphism $\Phi: \Bbb{Z}G_1\to \Bbb{Z}G_2$ such that for $\alpha= \sum_{i=1}^nr_ig_i \in \Bbb{Z}G_1$, $\Phi(\alpha)=\sum_{i=1}^{n}\varepsilon(\phi(g_i))^{-1}r_i\phi(g_i)$.
$\Phi$ is normalized in the sense that $\varepsilon(\Phi(\alpha))=\varepsilon(\alpha)$ .

Now as $G_1$ is finite abelian, $\Bbb{Z}G_1$ is commutative $\implies $ $G_2$ is abelian. Now as $\Bbb{Z}$ has IBN, it implies $|G_1|=|G_2|$.

Now as each $g \in G_1$ is invertible in $\Bbb{Z}G_1$, so $\Phi(g)$ is a normalized unit in $\Bbb{Z}G_2$, and it is not so hard to check that all units of integral group ring of an abelian group are trivial (easier if we have finite abelian, which is our case), so $\Phi(g)\in \pm G_2$ and $\Phi$ is normalized implies $\Phi(g)\in G_2$. This shows that $\Phi(G_1) \subset G_2$ but $|G_1|=|G_2|$, we get $\Phi(G_1)=G_2$. $\hspace{7cm} \blacksquare$

It is also true for finite metabelian groups.

Moreover, your hunch was correct that you can deduce something about enteries in character table, to be precise, by using a theorem by Glauberman, you can also prove that

if $G_1$ is finite such that $\Bbb{Z}G_1\cong \Bbb{Z}G_2$, then character tables of $G_1$ and $G_2$ are equal (after a possible rearrangement).

See Sehgal, Milies book " An Introduction to group rings "

You might consult Chapter 9 in Milies, Sehgal, "An Introduction to Group Rings". It discusses how several invariants of groups are encoded in their integral group rings, and how this leads to an affirmative answer of the integral group ring isomorphism problem for some classes of groups.