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][1] 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.


  [1]: http://en.wikipedia.org/wiki/HNN_extension#Britton.27s_Lemma