Motivated by this question, it seems natural to ask the following:

Question 1: Assume $G$ is a [finitely generated discrete] torsion-free virtually Abelian group. Let $R = \mathbb{F}_2[G]$ be its group ring over $\mathbb{F}_2$. Look at the multiplicative monoid of $R \setminus \{0\}$. Is this monoid [right-]amenable?

(Amenable here means the existence of an invariant mean i.e. an element $\mu$ in the dual of $\ell^\infty(M)$ so that $\mu$ is invariant under the translation action of the monoid; I don't actually care if it's right or left amenable, but it only needs to be one of both.)

The question whose status I would actually like to know (and which has a higher chance of attracting a negative answer) is :

Question 2: Assume $G$ is a [finitely generated discrete] torsion-free amenable group. Let $R = K[G]$ be the group ring for some finite field $K$ and assume the zero-divisor conjecture holds for $G$. Look at the multiplicative monoid of $R \setminus \{0\}$. Is this monoid [right-]amenable?

Lastly one could also replace the base ring $K$ by $\mathbb{Z}$. I stumbled on this question while trying to give a tentative [positive] answer over there, but it seems these questions should have been settled long ago. Or maybe the answer to this question (for some group) has an implication to an open problem? In any case, it would be nice to have the status of this (probably?) well-studied question.

This thesis gives a good overview of amenability for semigroups (and hence monoids). Note that if the group $G$ is Abelian, then the multiplicative monoid of $K[G]$ is also Abelian and hence amenable (result which goes back to Day or Markov/Kakutani).

PS: I'm hitting lots of pay-walls when going through references, so apologies for not being able to complete the history check.

EDITS: removed the zero from the monoid (otherwise the measure supported at 0 is a mean); added torsion-free hypothesis to avoid zero-divisors in Question 1; added that the zero-divisors should hold in Question 2 for the same reason.

Motivated by this question, it seems natural to ask the following:

Question 1: Is there a [finitely generated discrete] torsion-free virtually Abelian (but not Abelian) group $G$ so that the multiplicative monoid $R \setminus \{0\}$ of $R = \mathbb{F}_2[G]$ its group ring over $\mathbb{F}_2$ is [right-]amenable?

Note that for virtually Abelian groups the zero-divisor conjecture holds, so it makes sense to speak of the multiplicative monoid of $R$. Amenable here means the existence of an invariant mean i.e. an element $\mu$ in the dual of $\ell^\infty(M)$ so that $\mu$ is invariant under the translation action of the monoid; I don't actually care if it's right or left amenable, but it only needs to be one of both. If the group is Abelian, then the monoid is also Abelian hence amenable.

Question 2: Is there a [finitely generated discrete] torsion-free non-Abelian amenable group $G$ so that: (1) the zero-divisor conjecture holds for $R = K[G]$ (the group ring over some finite field $K$) and (2) the multiplicative monoid of $R \setminus \{0\}$ is [right-]amenable?

Lastly one could also replace the base ring $K$ by $\mathbb{Z}$. I stumbled on this question while trying to give a tentative [positive] answer over there, but this tentative answer is false and it seems the questions above should have been settled long ago. Or maybe the answer to this question (for some group) has an implication to an open problem? In any case, it would be nice to have the status of this (probably?) well-studied question.

This thesis gives a good overview of amenability for semigroups (and hence monoids). Note that if the group $G$ is Abelian, then the multiplicative monoid of $K[G]$ is also Abelian and hence amenable (result which goes back to Day or Markov/Kakutani).

PS: I'm hitting lots of pay-walls when going through references, so apologies for not being able to complete the history check.

First EDIT: removed the zero from the monoid (otherwise the measure supported at 0 is a mean); added torsion-free hypothesis to avoid zero-divisors in Question 1; added that the zero-divisors should hold in Question 2 for the same reason.

Second EDIT: in view of recent developments, I changed the formulation of the question to make them still pertinent.