If $G$ is a finite abelian group, then $\mathbb C[G] = \lbrace f \colon \hat G \to \mathbb C \rbrace$, where $\hat G$ is the Pontrjagin dual of $G$. The isomorphism $g \mapsto g^{-1}$ translates into the same map on the Pontrjagin dual (basically multiplication by $-1$ on $\hat G$), but now it is a bit easier to analyze. Note also, that there is a non-canonical isomorphism $G \cong \hat G$.
Hence, in order to find two non-isomorphic abelian groups which have strongly isomorphic complex group rings, we just have to analyze (in addition to the cardinality of the group) the orbit structure of multiplication by $-1$ on the group. Indeed, we are now just talking about an algebra of function on a set with some $\mathbb Z/2 \mathbb Z$-action.
Example: In $\mathbb Z/8\mathbb Z \times \mathbb Z/2\mathbb Z$, there are precisely $4$ elements, which are fixed under multiplication by $-1$, namely $(0,0),(4,0),(0,1)$ and $(4,1)$. The same is true for $\mathbb Z/4\mathbb Z \times \mathbb Z/4\mathbb Z$. Here, one has $(0,0),(2,0),(0,2)$ and $(2,2)$. Hence, there exists an isomorphism between $\mathbb C[\mathbb Z/8\mathbb Z \times \mathbb Z/2\mathbb Z]$ and $\mathbb C[\mathbb Z/4\mathbb Z \times \mathbb Z/4\mathbb Z]$, which respect the isomorphism which is induced by $g \mapsto g^{-1}$.
I do not know about an example with coefficients in $\mathbb Z$.