Let $X$ be a path connected space, and let $x \in X$. Then we have that $\pi_1 (X,x)$ is a full subcategory of $\Pi(X)$. So, the inclusion functor $J: \pi_1(X,x) \to \Pi(X)$ is an equivalence of categories.
May, in his proof of the Van Kampen Theorem, asserts the following.
An inverse equivalence $F: \Pi(X) \to \pi_1(X,x)$ is determined by a choice of path classes $x \to y$ for $y \in X$.
I'm not sure how. I think he means that for each $y \in X$, one chooses a path class $\alpha_y :x \to y$, and using this maps a morphism $f:y \to y'$ by $F$ as follows: $$Ff := \alpha_{y'}^{-1} \circ f \circ \alpha_y$$
Is this the case?
Having realized the answer, I'm answering my own question.
The approach in the original post is the correct one, with one slight addition; we must take $\alpha_x$ to be the identity.
The approach works in general for any groupoid, as long is it is connected. To see that the definition gives a functor, we simply see that for $g:y \to y'$ and $h: y' \to y''$, $F(h \circ g) = \alpha_{y''}^{-1} \circ h \circ g \circ \alpha_y = F(h) \circ F(g)$ (By inserting $\alpha_{y'}\circ\alpha_{y'}^{-1}$). Note that $F$ is fully faithful and surjective.
Also, we get that this is precisely an inverse of $J$, because $F \circ J$ is $\operatorname {id}_{\pi_1(X,\, x)}$ and $J \circ F$ is naturally naturally isomorphic to the identity on $\Pi(X)$.