I don't think that there is a generating set for which this is possible.
Let $P$ be the random walk operator, $P = I-L$.
Let $g_n = \sum_{i=0}^n P^n \delta_x$ where $\delta_x$ is the Dirac mass at some vertex.
Clearly $\| g_n\|_{\ell^1}= n+1$
and $L g_n = I - P^{n+1}$.
So $\| L g_n\|_{\ell^1} \leq 2$.
This shows that $\frac{ \|L g_n\|_{\ell^1}}{\|g_n\|_{\ell^1}} \leq \frac{2}{n+1}$ tends to 0. So 0 is in the spectrum and the Laplacian cannot be inverted.

Also, be careful: the image of the Laplacian is not dense.

To see this first write the Laplacian $L$ as the composition of two operators $\nabla$ (the gradient) and $\nabla^*$ the divergence. 

To define those, let $X$ is the set of vertices and $E \subset X \times X$ be the set of edges (edges are here oriented, one can also impose that $E$ is symmetric: $(x,y) \in E \implies (y,x) \in E$; this has no real influence on the problem). Given a function $f: X \to \mathbb{C}$, let $\nabla f(x,y) = f(y) - f(x)$ ($\nabla f$ is a function on the edges).

$\nabla^*$ is the adjoint of $\nabla$. Given a function $g$ on the edges, $\nabla^*g(x) = \sum_{y \sim x} g(y,x) -\sum_{y \sim x} g(x,y)$ where $y \sim x$ are the neighbours of $x$ ($\nabla^*g$ is a function on the vertices).
Easy computations show that $L = \frac{1}{2d} \nabla^* \nabla$ where $d$ is the degree of the vertices.

$\mathrm{Im} \nabla^*$ is $\ell^1_0X = \lbrace f \in \ell^1X \mid \sum_{x \in X} f(x) =0 \rbrace$. So the image of $L$ is contained in $\ell^1_0X$. 
One can further check that $\overline{\mathrm{Im} L} = \ell^1_0 X$ if and only if the graph has no non-constant bounded harmonic functions.

Non-amenable groups always have non-constant bounded harmonic functions so that the closure of the image of $L$ is strictly contained in $\ell^1_0 X$.