Let $G$ be a finitely generated group (optionally torsion-free). Let $N$ be a submonoid of $G$ (that is, a subsemigroup with $1$). 


A (cancellative) monoid/semigroup $S$ is *right reversible* if for all $a,b \in S$, it follows that $Sa \cap Sb \neq\emptyset$. This is the Ore condition for semigroups, and allows one to invert the elements of $S$ so that the set of products, $S\cdot S^{-1}$, is a group. 


Suppose that $G$ is nilpotent. Does it follow that every submonoid $N$ is right reversible?


What are the *right* conditions on $G$ to guarantee this property? 


Something like nilpotent by finite comes to mind, since this implies the group ring ${\bf Z} G$ is Goldie. Obviously, this is a strong condition on $G$, since it fails for large classes of groups, e.g., those containing a non-abelian free group, or probably, those for which the group ring is not Goldie.


The question arose from a study of random walks on discrete groups, related to the boundary of one of the usual boundaries. It is a little too complicated to explain the connections here. 



Right reversibility of submonoids of nilpotent groups
It follows from Lemma 2 in Grigorchuk's <https://page-one.live.cf.public.springer.com/pdf/preview/10.1007/BF01138837> that all subsemigroups of a group $G$ are right reversible iff it contains no free semigroup on 2 generators. 


A free subsemigroup is not right reversible. Conversely suppose $S$ is not right reversible (say $aS\cap bS=\emptyset$, note for me right reversible is dual to the OP's definition). Then observe that $a,b$ generate a free subsemigroup. For words beginning with distinct letters are not equal and if they begin with the same letter you can cancel it to get a shorter relation. 


