Wedge product $S^1 \vee S^2$

It's correct, even intuition suggest us that in this case the only kind of paths that aren't in the same equivalence class of the trivial one is the paths around $S^1$. And moreover it is not important that they move around the sphere because I can homotopically retract them to move only around the circle if they move around it.

if fact you have just applied a more general corollary of Van Kampfen which is

$\pi_1(X \vee Y) $ is the group with generators the union of generators of $\pi_1(X)$ and $\pi_1(Y)$ and with relations the union of relations of the two groups

the proof is easy, just choose the two opens $X$, $Y$ and intersection a simply connected neighborhood of the wedge point