Igor, I think it is still (mostly) unknown. Suppose that $\Gamma$ is a product of non-uniform irreducible lattices $\Gamma_i$. If all factors $\Gamma_i$ are lattices in rank 1 Lie groups then quasi-isometries preserve the product structure according to our paper
 Kapovich, Kleiner, Leeb, Quasi-isometries and the de Rham decomposition,
Topology 37 (1998), no. 6, 1193–1211.
The reason is that in this case each $\Gamma_i$ contains quasi-geodesics with exponential divergence, so it is of Type I in the sense of . Once you know this, you are in business because the factors $\Gamma_i$ are QI rigid. However, if you allow factors which are non-uniform lattices of rank $\ge 2$, then, conjecturally, they have linear divergence. Special cases of this conjecture are proven in
 Drutu, Mozes, Sapir, Divergence in lattices in semisimple Lie groups and graphs of groups. Trans. Amer. Math. Soc. 362 (2010), no. 5, 2451–2505.
Thus, such non-uniform lattices (at least conjecturally) are of neither type I nor II (in the sense of ), so  does not apply and, at this point (I think) no other technique is available to handle quasi-isometries of products. However, you should check with Kevin Wortman, since in his work on S-arithmetic lattices and lattices in algebraic groups over functional fields he had to handle similar issues. Thus, there is a chance that QI rigidity for reducible lattices is implicit in his work.
Another possible approach would be to generalize  using the fact that "higher-dimensional" exponential divergence is now known for non-uniform lattices of higher rank.