I think that it is possible to give a positive answer to this, in a slightly larger level of generality.

Let me start by giving the hypotheses that we shall work with (the arguments work in an even more general setting, but I think that the following is reasonable for the purposes here).

**Context 1** Let $k$ be a field, and let $G$ be an affine group scheme over $k$. Let $X$ be a finite type $k$-scheme with a $G$-action, and suppose that there is a morphism $\pi: X \to Y$ satisfying the conditions 1-3 of the post. Furthermore, we assume that the $G$-stabilizer of any geometric point of $X$ is trivial.

Just for completeness, the conditions 1-3 that we are assuming are:

(1) $\pi$ is affine, surjective, and $G$-invariant.

(2) We have $\mathcal{O}_Y = (\pi_*\mathcal{O}_X)^{G}$.

(3) $G \times X \to X \times X$ induces a surjective morphism $G \times X \to X \times_Y X$.

The following condition will be useful:

**Definition** We say that the action of $G$ on $X$ is proper if the induced morphism $G \times X \to X \times X$ is proper.

With this in mind, the following is not too difficult.

**Proposition 1** In Context 1, assume that the action is proper. Then, the morphism $\pi: X \to Y$ is a principal $G$-bundle.

*Proof:* The condition on the stabilizers means that the quotient stack $X/G$ is actually an algebraic space; equivalently $G \times X \to X\times X$ is a monomorphism. By definition, the quotient morphism $X \to X/G$ is a principal $G$-bundle. There is an induced morphism $f: X/G \to Y$; it suffices to show that $f$ is an isomorphism. This can be checked Zariski locally on $Y$, so we may assume that $Y$ and $X$ are affine for concreteness. The assumption of that the action is proper is equivalent to $X/G$ being separated, and so in particular the morphism $f: X/G \to Y$ is separated. On the other hand, (3) implies that $f$ induces an injection on geometric points, and so $f$ is separated and quasi-finite. By https://stacks.math.columbia.edu/tag/03XX, it follows that $X/G$ is a scheme. Notice that $H^0(\mathcal{O}_{X/G}) = H^0(\mathcal{O}_X)^G = H^0(\mathcal{O}_Y)$, where the last equality is by (2). Therefore $f_*(\mathcal{O}_{X/G}) = \mathcal{O}_Y$. By Zariski's main theorem https://stacks.math.columbia.edu/tag/02LR, it follows that $f$ is an open immersion. The surjectivity in (1) implies that the open immersion $X/G \to Y$ is surjective, and therefore $X/G \to Y$ is an isomorphism. QED

Now, I think(?) that the condition of properness of $G$ is automatic in Context 1. This seems quite strong, I am including an argument below, which we may have to double-check (as it seemed a bit surprising to me).

**Proposition 2 (please double-check)** In Context 1, the action is automatically proper. In particular, in Context 1 the morphism $\pi: X \to Y$ is always a principal $G$-bundle.

*Proof:* By the triviality of stabilizers we have that $G \times X \to X \times X$ is a monomorphism, and it factors through the closed subscheme $X \times_Y X \subset X \times X$, thus inducing a monomorphism $i: G \times X \to X \times_Y X$. It suffices to show that $i$ is a closed immersion, which by item (6) in https://stacks.math.columbia.edu/tag/04XV is equivalent to showing that $i$ is universally closed. The morphism $i: G\times X \to X \times_Y X$ is $G$-equivariant, where $G$ acts on the first coordinate of the source by multiplication, and it acts on the second coordinate of the target by the action on $X$. These actions are free, and taking quotients we get a monomorphism of algebraic spaces $\widetilde{i}: X \to X \times_Y (X/G)$. Note that $G \times X \to X$ and $X \times_Y X \to X \times_Y (X/G)$ are principal $G$-bundles, and by working flat locally on $X \times_Y (X/G)$ one can see that the following square is Cartesian:
$\require{AMScd}$
\begin{CD}
G \times X @>i>> X \times_Y X\\
@V V V @VV V\\
X @>>\widetilde{i}> X \times_Y (X/G)
\end{CD}
Therefore it suffices to show that $\widetilde{i}$ is universally closed. Notice that by (3) the morphism $i$ is a surjective monomorphism, and it follows that $\widetilde{i}$ is also a surjective monomorphism, in particular it is universally bijective on points. Furthermore, the morphism $\widetilde{i}$ has a section $p: X \times_Y (X/G) \to X$ given by the first projection. To check that $\widetilde{i}$ is universally closed, choose a morphism $T \to X \times_Y (X/G)$ from a scheme $T$ and form the base-change $\widetilde{i}_T: X_T \to T$; we need to show that $\widetilde{i}_T$ is closed. But it is still the case that $\widetilde{i}_T$ is a surjective monomorphism (so it is bijective on topological points) and has a section $p_T: T \to X_T$. Using this, we see that, for given closed subset $Z \subset |X_T|$of the topological space, the image $\widetilde{i}_T(Z)$ coincides with the preimage $(p_T)^{-1}(Z)$, and so it is closed, as desired. QED