Morita equivalent rings have the same center (exercise). The center of a group ring $k[G]$ has a basis given by sums over the conjugacy classes of $G$, and in particular has dimension given by the number of conjugacy classes of $G$ (exercise). A finite group of size at least $2$ has at least $2$ conjugacy classes.
(Edit: As pointed out by Eric Wofsey in the comments, this argument only shows that the centers are not isomorphic as $k$-algebras, so only addresses Morita equivalence over $k$.)
If $G$ is infinite then the center of $k[G]$ involves only sums over the finite conjugacy classes, and so can be trivial if every nontrivial conjugacy class of $G$ is infinite. So we can instead argue as follows. I assume $k$ is a field, in which case the only rings Morita equivalent to $k$ are $M_n(k)$, which are in particular all finite-dimensional over $k$. You can also use this to finish: there's an obvious homomorphism $k[G] \to k$ given by sending all $g \in G$ to $1$, and $M_n(k)$ doesn't admit such a homomorphism unless $n = 1$.