There is no generating set for which this is possible (actually this is true for any graph with an infinite connected component).

The functions $g_n$ which show 0 are in the spectrum are constructed by trying to invert the Laplacian naively

*Remark*: My convention for the Laplacian seem to have the opposite sign as yours, so I'm using $\Delta = -L$

Let $P$ be the random walk operator, $P = I- \Delta$ or $P = \frac{1}{|S|} \sum_{g \in S} g$.

Let $g_n = \sum_{i=0}^n P^n \delta_x$ where $\delta_x$ is a Dirac mass at some vertex $x$.
Since $P^i \delta_x$ is the probability distribution of the random walk at time $i$, $\|P^i\delta_x\|_{\ell^1}$. These are positive functions, so their norm add up in $\ell^1$: hence $\| g_n\|_{\ell^1}= n+1$

On the other hand, $\Delta = I-P$ so and $\Delta g_n = \delta_x - P^{n+1} \delta_x$.
So $\| \Delta g_n\|_{\ell^1} \leq \|\delta_x\|_{\ell^1} + \| P^{n+1} \delta_x\|_{\ell^1} = 2$ by the triangle inequality.

This shows that $\displaystyle \frac{ \|\Delta g_n\|_{\ell^1}}{\|g_n\|_{\ell^1}} \leq \frac{2}{n+1}$ which obviously tends to 0. 
So 0 is in the spectrum and the Laplacian cannot be inverted $\ell^1$.

Note the argument is true in any graph (without finite connected components).

Also **the image of the Laplacian** is not dense (it is weak$^*$ dense as the sequence $g_n$ above shows).

It's easy to check that its image lies in $\ell^1_0X = \lbrace f \in \ell^1X \mid \sum_{x \in X} f(x) =0 \rbrace$. 

One can further check that $\overline{\mathrm{Im} \Delta} = \ell^1_0 X$ 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 $\Delta$ is strict subspace of $\ell^1_0 X$.