The Laplacian is never invertible on $\ell^1(G)$.
There is already a correct answer by ARG. I am writing this answer as there is a demand for more information. I will thus try to provide a more conceptual POV. But I should remark that my answer is essentially the same as ARG's answer.
Let $\mu$ be a probability measure on $G$ and assume the Laplacian $\Delta=1-\mu$ is an automorphism of $\ell^1(G)$. By taking the double dual, $\Delta$ extends to an automorphism of $\ell^\infty(G)^*$. Recall that a mean on $G$ is a norm 1 positive element of $\ell^\infty(G)^*$ and let $M$ be the space of all means on $G$. It is easy to see that $M$ is convex and compact for the w*-topology. Note that $G$ acts on $M$ preserving its structure. In particular, $M$ is $\mu$-invariant. By Markov–Kakutani fixed-point theorem there is a $\mu$-invariant mean $m\in M$. Thus $\ker \Delta\neq 0$ as $$ \Delta m=m-\mu m =0. $$ This is a contradiction.