I am interested in the following question: Can semisimple Lie group of real rank $\geq 2$ contain an abelian lattice?
No, and it has nothing to do with the higher-rank assumption.
Probably the easiest way to see this is by the Borel Density Theorem, which says that if $G$ is a semisimple Lie group which has no compact factor and it is algebraic, and if $\Gamma$ is a lattice in $G$, then $\Gamma$ is Zariski dense in $G$.
Now, if you have a non-compact semisimple Lie group and a lattice in it, you can mod up the center and get an algebraic group and a lattice in it. Further, the algebraic group is a product of factors and by modding out compact ones you are in the situation described above.
Another easy way to see that a lattice cannot be abelian is by showing that it cannot be amenable. If it was then the enveloping group would be amenable as well, but it is not. Eg it contains a closed free group.
No. A lattice in a semisimple Lie group of higher rank has to contain distorted elements of infinite order (a theorem of Lubotzky, Mozes and Raghunathan). On the other hand, every element of infinite order in a finitely generated abelian group is undistorted.
Edit: I missed the assumption that the lattice should be non-uniform (Thanks Yves).