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.

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.

FURTHER EDIT:

Here are some more details, to exacerbate Tom's delight/revulsion.  The Milnor--Wolf Theorem asserts that every solvable group is either virtually nilpotent (ie has a nilpotent finite-index subgroup) or contains a free subsemigroup.  (Googling "Milnor--Wolf Theorm" gives abundant references.)  S is a solvable group that is not virtually nilpotent, so it contains a free sub-semigroup, which may as well be of rank two.  Let T be the group generated by this sub-semigroup.

Now S decomposes as a group extension -

1->Z^2->S->Z->1

and T inherits a decomposition as an extension

1->(Z^2\cap T)->T->T'->1.

Because T isn't abelian, T' is non-trivial.  If Z^2\cap T is cyclic or trivial then T is nilpotent, which contradicts the fact that T contains a free semigroup.  Therefore Z^2\cap T is finite-index in Z^2 and so T contains a copy of Z^2.

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