I don't think Richard's idea is quite right. If G is the kernel of the map to Z^2 then ab and ba lie in the same coset of G, so G can't have property (P).
But there is a simple solvable example. Consider the standard example of a lattice in Sol, namely the semidirect product of Z^2 (generated by a,b ) by Z (generated by t) in which tat^{-1}=2a+b and tbt^{-1}=a+b. This group, S, is generated by a and t and contains a rank-two abelian subgroup.
You can think of S as an HNN extension. Britton's Lemma implies that any word w in t and a is reducible if and only if it 'obviously' is, ie if and only if you see something of the form tat^-1 or t^-1at. In particular, every positive word in a and t is reduced and so a and t generate a free semigroup.
EDIT:
Greg points out that this doesn't work. But it can be fixed with a nasty hack. By the Milnor--Wolf Theorem, the group S does contain a non-abelian free semigroup, which we may as well take to be two-generator. Let T be the subgroup generated by these two generators. Now it's not hard to convince yourself that the only possibility is that T is a finite-index subgroup of S. (Otherwise, T would be nilpotent, and so have polynomial growth.) So T is generated by two elements that generate a free semigroup, and contains a copy of Z^2.