Does there exist a subspace of $\Bbb R^2$ with fundamental group $\Bbb Z \oplus \Bbb Z$?
No. I prove in this answer that $H_2(X;\Bbb Z)$ surjects onto $H_2(K(\pi_1 X, 1);\Bbb Z)$. (Actually, the coefficients can be whatever you want.) If you want to do this for gross $X$ this is still true but you need to use CW approximation. Further, for $X$ compact, it follows immediately from (Cech) Alexander duality that $H_2(X) =0$. If you further assume that $X$ is locally contractible, you can drop the Cech parenthetical; in this case, the techniques used in the proof can all be found in a first-year course on algebraic topology (even though they require heavier machinery than the fundamental group alone).
Unfortunately without these assumptions this gets more difficult. It is apparently a theorem of Zaslow that the higher homology of any subset of the plane vanishes identically. For $\pi_1 X = \Bbb Z^2$, $K(\Bbb Z^2,1) = T^2$, so we see that the second homology of any space with $\Bbb Z^2$ fundamental group is nontrivial.
In fact, being a subset of the plane places severe restrictions on the fundamental group; see here. In particular, any finitely generated subgroup is free. I consider this much harder than the first paragraph, but that's probably partly because I only know how to prove what's in the first paragraph.