Let $G$ be an abelian group and let $R$ be a commutative ring and consider the group ring $R[G]$ of finite formal linear combinations of elements of $G$ with coefficients in $R$. Let $J = (1 - g ~|~ g \in G)$ be the augmentation ideal. If $R[G]$ is semisimple, then $J = J^2$. One way to see this is to consider the short exact sequence of $R[G]$-modules \begin{equation*} 0 \to J/J^2 \to R[G]/J^2 \to R[G]/J \to 0. \end{equation*} If $R[G]$ is semisimple then this splits, which implies that $J / J^2 = 0$.
Is the converse true?
If $R = K$ is a field and $G$ is finite, I believe the answer is yes: $J / J^2 = K \otimes_\mathbb{Z} G$, which is nonzero exactly when the characteristic of $K$ divides the order of $G$, i.e. exactly when $K[G]$ is not semisimple.
I think (maybe?) that it's always true that $J/J^2 = R \otimes_\mathbb{Z} G$, but then I worry about whether or not the rest of the previous argument still works. For me, that argument is an argument about representations, which get very nasty very quickly when $R$ is not a field.