Reduction from 2-SAT to 2-XOR-SAT? 2-SAT is unsatisfiable iff it contains unsatisfiable XOR-2-SAT. 

So, first, we just need to combine every clause that contains share same variables (both of them). Then, all of those which remain uncombined are "uninteresting" (don't affect satisfiability).

Example:

$\Phi = (x \lor y) \land (\overline x \lor \overline y) \land (x \lor \overline z)\land(\overline x \lor z)\land (\overline y \lor z)\land (y \lor \overline z)\land (z \lor t)$

Now I can replace this formula by equisatisfiable (under log-space reduction):

$\Phi' = (x \oplus y)\land(x \oplus \overline z)\land(y \oplus \overline z)$

As you can see, clause $(z\lor t)$ was removed as "uninteresting".

Ah, well, I need to see if there are cycles in implication graph. So, finding a cycle in directed graph that contains two given vertices is $\mathsf{NL}$-complete problem, right?
This method doesn't work since you can't always pair clauses as in your example.

According to Wikipedia, 2SAT is NL-complete. In particular, this means that it is not in logspace unless L=NL, which is considered unlikely.