Existence of Folner Sequence Implies Left Invariant Mean

I don't really understand what $A$ is supposed to be in your attempt. However, let me explain what I think is a natural approach here.

The following is a classical exercise in functional analysis (the proof uses Hahn-Banach)

Exercise: Show that there exist a Banach limit on $\ell^\infty$, i.e. a bounded linear functional of norm one, namely $\phi\in(\ell^\infty)^*$ with the following properties:

1. if $\ell^\infty\ni x\ge0$, i.e. $x=(x_n)$ with $x_n\ge0$, then $\phi(x)\ge0$
2. If $x=(x_n)\in\ell^\infty$ is convergent, then $\phi(x)=\lim_{n\to\infty}x_n$.
3. If $x=(x_n)\in\ell^\infty$, then $\phi(x_1,x_2,x_3,\dots)=\phi(x_2,x_3,\dots)$ (the sequence where we just ignored the 1st term of $x$).

Now suppose that you have solved the exercise; I do not include a proof as there are tons of proof online, just search for existence of Banach limits if you cannot solve it yourself (I would advise you to try it, it's a fun exercise; here is a hint though: you might need to use this consequence of Hahn-Banach). You can get your invariant mean for free: define $\lambda:\ell^\infty(G)\to\mathbb{R}$ by $$\lambda(f)=\phi\bigg(\bigg\{\frac{1}{|A_n|}\sum_{g\in A_n}f(g)\bigg\}_{n=1}^\infty\bigg)$$

It is trivially verified that $\lambda$ is a bounded linear functional, $\lambda(1)=1$ and $f\ge0\implies\lambda(f)\ge0$. So all one needs to verify is that $\lambda$ is translation invariant. For that we will use property (3) of $\phi$.

Fix $s\in G$ and $f\in\ell^\infty(G)$. For any $n\ge1$, do the estimate: $$\bigg|\frac{1}{|A_n|}\sum_{g\in A_n}f(g)-\frac{1}{|A_n|}\sum_{g\in A_n}\tau_s(f)(g)\bigg|=\frac{1}{|A_n|}\bigg|\sum_{g\in A_n}f(g)-\sum_{g\in s^{-1}A_n}f(g)\bigg|\le\frac{|A_n\triangle s^{-1}A_n|}{|A_n|}\|f\|_\infty $$ (convince yourself that the above estimate is true). Now given $\varepsilon>0$, find $N$ large enough so that, for all $n\ge N$ we have $$\frac{|A_n\triangle s^{-1}A_n|}{|A_n|}\|f\|_\infty<\varepsilon$$ And, finally, by iterating property (3) of $\phi$ $N$ times, you get that $$\lambda(f)=\phi\bigg(\bigg\{\frac{1}{|A_n|}\sum_{g\in A_n}f(g)\bigg\}_{n=N}^\infty\bigg)$$ and that $$\lambda(\tau_s(f))=\phi\bigg(\bigg\{\frac{1}{|A_n|}\sum_{g\in A_n}\tau_s(f)(g)\bigg\}_{n=N}^\infty\bigg)$$ and thus $$|\lambda(f)-\lambda(\tau_s(f))|\leq\sup_{n\ge N}\frac{|A_n\triangle s^{-1}A_n|}{|A_n|}\|f\|_\infty<\varepsilon$$ As $\varepsilon$ is arbitrary, allowing it to approach $0$ yields $\lambda(f)=\lambda(\tau_s(f))$.