From $P=NP$ to $NP=NL$
Yes and no.

If P=NP, then every problem in NP, in particular 3SAT, can be solved in polynomial time. Hence it reduces to every nontrivial problem (and 2SAT is nontrivial) using polynomial time reductions. Which are the reductions in the definition of NP-hardness.

NL, on the other hand, is not closed under polynomial time reductions (unless NL=P), as these can use polynomial space (exceeding logarithmic space). That is why for the definition of NL-hardness, logspace reductions are used. Hence neither can we conclude that 3SAT is in NL, nor that NL=NP.