Here is (I believe) a simpler argument, combining 1--6 into one step. (I assume here the group is countable; I can't tell if you're interested in uncountable discrete groups, but I have no idea what issues, if any, arise there.)

Let $G$ be a countable abelian group generated by $x_1,x_2,\ldots$. Then a Følner sequence is given by taking $S_n$ to be the pyramid consisting of elements which can be written as

$a_1x_2+a_2x_2+\cdots+a_nx_n$ with $|a_1|\leq n,|a_2|\leq n-1,\ldots,|a_n|\leq 1$.

The invariant probability measure is then defined by $\mu(A)=\underset{\omega}{\lim}|A\cap S_n| / |S_n|$ as usual.

A more natural way to phrase this argument is:

1. The countable group $\mathbb{Z}^\infty$ is amenable.
2. All countable abelian groups are amenable, because amenability descends to quotients.

But I would like to emphasize that there is really only one step here, because the proof for $\mathbb{Z}^\infty$ automatically applies to any countable abelian group. This two-step approach is easier to remember, though. (The ideas here are the same as in my other answer, but I think this formulation is much cleaner.)