Let me describe how I would try amenability. I'll make it in two steps (the first step is fine; the second is more of a work in progress) by going through the [possible] amenability of the monoid $R[G] \setminus \{0\}$. I discuss below why I think this strategy seems fine.

The first step is relatively straightforward.
Let me call a partially-absorbing element of a monoid $M$ an element $x$ such that for some $m \in M$ with $m \neq 1$, $xm = x$. Note that if the ring $R[\Gamma]$ has no zero divisors, there cannot be such an element except 0 (since $xm = x$ implies $x(m-1) =0$.

**Lemma 1:** let $M$ be an [right-]amenable monoid without partially-absorbing elements and $G$ be a subgroup of $M$. Then $G$ is amenable.

*Proof:* Let $\mu$ be a right-invariant mean on $\ell^\infty(M)$, that is $\forall m \in M, \forall f \in \ell^\infty(M)$ one has $\mu(\rho_mf) = \mu(f)$ (here $\rho_mf(x) := f(xm)$). Now consider the right-$G$-cosets of $M$, and choose representatives $m_i$ so that $\{m_iG\}$ covers $M$. (Note that being in the same $G$-coset is an equivalence relation because $G$ is a group; also note that the cosets are "naturally" identified with $G$ since there are no partially-absorbing elements).

Define a mean $\mu_G$ on $G$ as follows. For $f\in \ell^\infty(G)$ consider $f_M \in \ell^\infty(M)$ by $f_M(m) = f(g)$ where $m \in Gm_i$ so that $m = m_ig$. (Basically $f_M$ is a copy of $f$ on each $G$-coset.)
Define $\mu_G(f) = \mu(f_M)$.
Note that $\rho_gf_M(x) = f_M(xg)$.
Then $\mu_G(\rho_gf) = \mu(\rho_gf_M) = \mu(f_M) = \mu_G(f)$.
From there $G$ has an invariant mean and is consequently amenable (on the right, but then also on the left)$\qquad \square$

**Remark:** It follows from Lemma 1 that if $R[\Gamma] \setminus \{0\}$ is amenable [and satisfies the zero divisor conjecture] then $\Gamma$ is amenable (because $\Gamma$ is embedded in $R[\Gamma] \setminus \{0\}$ by sending $\gamma$ to the element solely supported on $\gamma$ and with coefficient 1.

Now the next "step" (the amenability of $R[G] \setminus \{0\}$) should be well-known (or known to be open? or perhaps well-known to be false? ).
I posted it as a separate question.

Before moving on, let me just point out why this strategy does not seem so bad.
Start with $\Sigma_0 =\Gamma$ (embedded as Dirac masses), then consider the monoid $\Sigma_1$ which is generated by elemetns of the form $1+\gamma$, then consider the monoid $\Sigma_2$ which is generated by elements with support of size $\leq 3$, etc.
This gives a step by step process which seems easier to tackle (either in postive [amenable] or negative [non-amenable]).
One might think that this brings too many elements, but the conjugate of the unit element $p$ under $\Gamma$ has an infinite orbit.
So it seems likely that any positive [or negative] argument for the monoid will transfer to the group of units.
The obvious argument against this strategy, is that there are many more tools to conclude of the amenability of groups, than of monoids.

**Problem 2:** Assume $\Gamma$ is a countable amenable group, $R$ is a finite field and that the group ring over $R[\Gamma]$ satisfies the zero divisor conjecture. Show that the multiplicative monoid $M = R[\Gamma] \setminus \{0\}$ is a [right-]amenable monoid.

Applying a positive answer to Problem 2 would show that $R[\Gamma] \setminus \{0\}$ is an amenable monoid, and then applying Lemma 1 to the group of units of $R[\Gamma]$ shows it is an amenable group.

[**EDIT:** as updated in Gilles' answer above, the group of unit is not amenable. So the interesting problem would either be: 1- is there any non-abelian group for which the multiplicative monoid (without 0) is amenable? 2- which $\Sigma_i$ is the first to be non-amenable?]

*In support of* the amenability of $\Sigma_1$ (the monoid generated by the Dirac masses and the elements of the from $1+\gamma$).
Each element of $\Sigma_1$ can be written in the form $\gamma \prod_{\eta \in S} (1+ \eta)$ where $S \in (\Gamma)^N$ (is an ordered $N$-tuple).
The element $\prod_{\eta \in S} (1+ \eta)$ is closely related to $\prod_{\eta \in S} \eta$.
Indeed it will be (at most) supported on all the possible partial products of $\prod_{\eta \in S} \eta$.
So this second part of the semigroup is very close to $\Gamma$ itself.
The difference with $\Gamma$ can be "measured" by how the different ways of writing an element $h$ as a product so that the partial products are distinct.
In some sense, this makes it sound that groups which are far from being Abelian should have a non-amenable $\Sigma_1$.
The question is then whether virtually-Abelian groups are far enough from being Abelian.

*Note:* Bartholdi has a paper about amenable $K$-algebras, but the notion of amenability he uses does not match with the amenability of the multiplicative monoid $K[G] \setminus \{0\}$. He uses the sequences $F_i$ (which clearly don't work here) in his paper.

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