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

[1] 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 [1]. 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

[2] 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 [1]), so [1] 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 [1] using the fact that "higher-dimensional" exponential divergence is now known for non-uniform lattices of higher rank.