The answer is yes for any generating set. But be careful, the Laplacian is not surjective (see below).

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 $L = \nabla^* \nabla$.

One needs to show that there are no sequence of functions $g_n$ so that $\nabla g_n$ tends to $\ker \nabla^*$.
Non-amenability is equivalent to $\nabla: \ell^1X \to \ell^1 E$ having closed image (nowadays it a pretty common trick, see for example Woess' book).
Since the image is closed, one only needs to show that $\mathrm{Im} \nabla \cap \ker \nabla^* = 0$.

It turns out that, in $\ell^2$, $\mathrm{Im} \nabla = \ker \nabla^*$ (this is a pretty standard result on adjoints of opertaors). 
Hence $\mathrm{Im} \nabla \cap \ker \nabla^* = 0$ in $\ell^2$.
If the intersection was non-trivial in $\ell^1$, it would also be non-trivial in $\ell^2$.
Hence the intersection is trivial and $L$ has a closed image.

**The image** of $L$ is however distinct from $\ell^1X$.
For starters $\mathrm{Im} \nabla^*$ 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 image of $L$ is strictly contained in $\ell^1_0 X$.

[NB: will try to update references at some point]