TITLE: Subgroups of a group generated by a free semigroup
QUESTION [8 upvotes]: Let $F$ be a free semigroup (say, $2$-generated) which is embedded in a group $G$, and suppose that $G$ (as a group) is generated by $F$. The most simple such situation would be when $G$ is a free group, but there are lot of groups, besides free ones, which could occur in this situation (for example, $G$ could be solvable). Is it possible that $G$ contains a subgroup isomorphic to $\mathbb{Z} \times \mathbb{Z}$ (the direct product of two copies of the infinite cyclic group $\mathbb{Z}$)?
Update: thanks to all the people for a very interesting discussion. Sorry that I cannot award a few "accepted answers", so the only one goes to Greg who supplied the most lucid and explicit example.

REPLY [3 votes]: It would be schizophrenic to try to slice this into comments on the various answers, so:
Brandon's condition (P) does not imply that no two elements of the semigroup lie in the same coset. Let F=F2; from Richard's and Henry's answers we have the example of G = [F,F], which has (P), while nevertheless we have ab=ba. The same issue showed up in Henry's first solution.
However, continuing the construction nevertheless we obtain the universal 2-step solvable group Γ=F/[[F,F],[F,F]]. Since we know from Greg's and Henry's answers that free semigroups exist in 2-generated 2-step solvable groups, it follows from universality that the semigroup in Γ generated by a and b is free. Since Γ contains [F,F]/[[F,F],[F,F]], an infinite-rank free abelian group, we see that Brandon's construction does work, at least in this case.
(Please direct your upvotes to the people who came up with these examples, not me.)