fundamental groups of open subsets of the plane

The new question is very different from the original one. The answer to it is that every noncompact connected surface is homotopy-equivalent to a bouquet of circles, see proofs and references here. In particular, fundamental group is free and, hence, torsion-free.

Thus, the answer to the last question is: Yes, there is a description, namely they all are disjoint unions of bouquets of circles.

A (quite a bit) more difficult theorem is that the fundamental group of every open connected subset of $R^3$ is still torsion-free. In dimensions $\ge 4$ this is, of course, false.

I think it is still possible to have $2$-torsion. The special orthogonal group $SO(n)$ is orientable, yet has a fundamental group of $\mathbb{Z}/2\mathbb{Z}$ for all $n\geq 3$. To see that $SO(n)$ is orientable, note that $SO(n)$ has a Lie group structure, so its tangent bundle is trivial. Hence, the top exterior power of the cotangent bundle is trivial, meaning that $SO(n)$ has a non-vanishing form of top-degree.