There is no generating set for which this is possible (actually this is true for any graph with an infinite connected component). Furthermore, your claim that (in non-amenable groups) such $f_n$ cannot exist if $f_n \geq 0$ is false.
Here is a detailed proof.

**Lemma 1:** If $0 \leq v,w \in \ell^1G$ then $\|v+w\|_{\ell^1} = \|v \|_{\ell^1} + \|w\|_{\ell^1}$.

*Proof:* for all $g \in G$, $|v(g) + w(g)| = v(g) + w(g) = |v(g)| + |w(g)|$ this passes to the $\ell^1$-norm (by summing both sides over all $g$). $\hspace{10.5cm} \square$

**Lemma 2:** For any function $v \in \ell^1G$ and for any $g \in G$, $\|gv\|_{\ell^1} = \|v\|_{\ell^1}$. If further $v \geq 0$, $gv \geq 0$.

*Proof:* $gv$ has the same values as $v$ but permuted by the group action, so the norm is the same. The same reasoning shows that $Pv$ is positive when $v$ is. $\hspace{7.5cm} \square$

**Definition:** Let $P$ be the random walk operator: $P = \frac{1}{|S|} \sum_{g \in S} g$ or $P = L+I$ (where $I$ is the identity).

**Lemma 3:** for any $0 \leq v \in \ell^1G$, $\|Pv\|_{\ell^1} = \|v\|_{\ell^1}$ and $Pv \geq 0$.

*Proof:* Note that for any $g \in G$, $gv \geq 0$. $$
\begin{array}{rll}
\|Pv\|_{\ell^1}
&= \| \tfrac{1}{|S|} \sum_{g \in S} gv \|_{\ell^1}
& = \tfrac{1}{|S|} \| \sum_{g \in S} gv \|_{\ell^1} \\
&\overset{Lemma 1}{=}
\tfrac{1}{|S|} \sum_{g \in S} \| gv \|_{\ell^1}
&\overset{Lemma 2}{=}
\tfrac{1}{|S|} \sum_{g \in S} \| v \|_{\ell^1} \\
&= \| v \|_{\ell^1}
\end{array}
$$
$Pv$ is a strictly positive scalar times a sum of positive function, so it is positive. $\hspace{3cm} \square$

**Notation:** $P^k$ is $P$ applied $k$ times (e.g. $P^3 v= PPPv$) and $P^0 = I$.

**Lemma 4:** Let $f_n = \displaystyle \sum_{i=0}^n P^i v$ where $v \geq 0$ and $\|v\|_{\ell^1} =1$.
Then $\|f_n\|_{\ell^1}= n+1$.

*Proof:*
$$
\begin{array}{rll}
\|f_n\|_{\ell^1}
&= \displaystyle \| \sum_{i=0}^n P^i v\|_{\ell^1}
& \overset{Lemma 1}{=} \displaystyle \sum_{i=0}^n \| P^i v\|_{\ell^1}\\
& \overset{Lemma 3}{=} \displaystyle \sum_{i=0}^n \|v\|_{\ell^1}
& = n+1. \\
\end{array}
$$

**Lemma 5:** Let $f_n = \displaystyle \sum_{i=0}^n P^i v$ where $v \geq 0$ and $\|v\|_{\ell^1} =1$.
Then $\displaystyle \frac{\|L f_n\|_{\ell^1}}{\| f_n \|_{\ell^1}} \to 0$.

*Proof:* Let's compute $Lf_n$ using $L=P-I$:
$$
Lf_n = \displaystyle \sum_{i=0}^n LP^i v
= \displaystyle \sum_{i=0}^n (P-I)P^i v
= \displaystyle \sum_{i=0}^n \big( P^{i+1}v-P^i v \big)
= P^{n+1}v -v
$$
Now $\|Lf_n\|_{\ell^1} = \|P^{n+1}v -v\|_{\ell^1} \overset{TI}{\leq} \|P^{n+1}v\|_{\ell^1} + \|v\|_{\ell^1} = 2$ where $TI$ stands for the triangle inequality.

Using Lemma 4 $\displaystyle \frac{ \|L f_n\|_{\ell^1}}{\|f_n\|_{\ell^1}} \leq \frac{2}{n+1}$ which obviously tends to 0.

**Corollary:** 0 is in the spectrum of $L_1$ and the Laplacian cannot be inverted $\ell^1$.

Note the argument is true in any graph. If $v$ is supported on a finite component, then $\tfrac{1}{n+1} f_n$ tends to the constant function, which is indeed in the kernel. Otherwise $\tfrac{1}{n+1} f_n$ tends weak$^*$ to 0 but not in norm.

As a complementary remark **the image of the Laplacian** is not dense (in a graph with an infinite connected component).
By taking $v = \delta_x$, the Dirac mass at some vertex $x$, the sequence $f_n$ above shows its image is weak$^*$ dense.
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 (norm) closure of the image of $\Delta$ is strict subspace of $\ell^1_0 X$.