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 point of view. 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.
Note that, by the same argument, $\Delta$ is never an automorphism of $\ell^\infty(G)$. For completeness let me add that for $1<p<\infty$ and $\mu$ with a generating support, $\Delta$ is an automorphism of $\ell^p(G)$ iff $G$ is not amenable. For this recall that the Mazur map $f \mapsto \text{sgn}(f)|f|^{2/p}$ is a uniformly continuous homeomorphism from the unit sphere of $\ell^2(G)$ to the unit sphere of $\ell^p(G)$. It follows that there exist almost invariant vectors in $\ell^p(G)$ iff there exist almost invariant vectors in $\ell^2(G)$, and this happens iff $G$ is amenable (by Kesten). Thus, if $G$ is amenable $\Delta$ is not an automorphism (and this works also for $p=1$). If $G$ is not amenable, by the uniform convexity of the unit ball of $\ell^p(G)$ (which fails for $p=1$), we get that $\|\mu\|<1$, thus $(1-\mu)\sum_{n=0}^\infty \mu^n=1$ and $\Delta$ is indeed an automorphism.
