Let $G$ be a discrete group, and let $T:\ell^2(G)\to \ell^2(G)$ be a bounded linear operator which commutes with all right translations: that is, if $\xi\in \ell^2(G)$ and $g\in G$ then $T(\xi\cdot g) =T(\xi)\cdot g$. (In other words, $T$ belongs to the group von Neumann algebra.) Then if $T$ is surjective, it is invertible.
This follows from combining a result of Kaplansky with the fact that $C^\ast$-algebras are inverse-closed in containing $C^\ast$-algebras. My own feeling is that the result is tacit folklore but in any case it follows by duality from Theorem 3.2 in this paper.