Upon further thought, I think that may avoid the (very nontrivial) local structure theorems in my answer and even prove a more general result in characteristic $0$. I post this potential argument as a separate answer, since merging everything in one answer would be a bit bulky.
Here is the claim.
Statement. Let $S$ be a Noetherian scheme over $\mathbb{Q}$. Let $G \to S$ be a smooth group scheme, and let $X \to S$ be a smooth scheme. Suppose that $X$ is equipped with an action of $G$ such that the induced morphism $\alpha \times id: G \times_S X \to X \times_S X$ is an isomorphism of sheaves on the small etale site of $S$. Then, $X$ is a $G$-torsor.
To prove this, first we use the same claim as before. The proof is the same as in the other answer. I state here as two lemmas for completeness:
Lemma 1. Let $s \in S$ be a point with residue field $\kappa(s)$, and let $\overline{s}= \mathrm{Spec}(\overline{\kappa(s)}) \to S$ be the geometric point corresponding to a choice of algebraic closure of $\kappa(s)$. Then, the morphism $G_{\overline{s}} \times X_{\overline{s}} \to X_{\overline{s}} \times X_{\overline{s}}$ is surjective at the level of $\overline{s}$-points.
Proof of Lemma 1. Choose an $\overline{s}$-point $y$ of $X_{\overline{s}} \times X_{\overline{s}}$. It is defined over a finite separable extension of the ground field $\kappa(s)$. By smoothness of $X\times X \to S$, we may find an étale $S$-scheme $T \to S$ equipped with a lift $\overline{s} \to T$ and a point $\widetilde{y}: T \to X \times X$ such that $\overline{s} \to T \to X \times X$ is the point $y$ we started with. Now, by assumption the morphism on $T$-points $G(T) \times X(T) \to X(T) \times X(T)$ is surjective, and so there is a $T$-point $ \widetilde{z}: T \to G \times X$ that maps to $\widetilde{y}$. The composition $\overline{s} \to T \xrightarrow{\widetilde{z}} G \times X$ maps to the original $y$, and we conclude the desired surjectivity. QED
Lemma 2. For all geometric points $\overline{s} \to S$, the fiber $X_{\overline{s}}$ is $G_{\overline{s}}$-equivariantly isomorphic to $G_{\overline{s}}/H$ for some closed subgroup $H \subset G_{\overline{s}}$.
Proof of Lemma 2. We may suppose that $\overline{s}$ is the algebraic closure of the residue field $\kappa(s)$ of its image $s \in S$. Pick a point $x \in X_{\overline{s}}(\overline{s})$. Set $H = \operatorname{Stab}_{G_{\overline{s}}}(x)$. There is an induced morphism $f:G_{\overline{s}} / H \to X_{\overline{s}}$. It suffices to show that $f$ flat, and that the scheme-theoretic preimage of every $\overline{s}$-point is a reduced point. By transitivity of the action of $G_{\overline{s}}$ on $\overline{s}$-points and the definition of $H$, we see that the preimage of every closed point is a reduced point. On the other hand, we know that $X_{\overline{s}}$ is smooth, and by descent the source $G_{\overline{s}}/H$ is also smooth. By miracle flatness, we have that $G_{\overline{s}}/H \to X_{\overline{s}}$ is flat. QED
Lemma 3. The quotient stack $[X/G]$ is a gerbe over $S$.
Proof of Lemma 3. We first show that $[X/G]$ is a gerbe over an $S$-algebraic space $Y$. By https://stacks.math.columbia.edu/tag/06QJ, to show this it suffices to prove that the relative inertia of $[X/G] \to S$ is flat. This would follow from the morphism $G \times_S X \to X \times_S X$ being flat. By the fiberwise criterion for flatness https://stacks.math.columbia.edu/tag/039E, it suffices to check that for all $s \in S$, we have that the induced morphism on $s$-fibers $G_s \times X_s \to X_s \times X_s$ is flat. We may even check this after passing to geometric points $\overline{s} \to S$, and by Lemma 1 we are reduced to checking flatness of the corresponding morphism when $X_{\overline{s}} = G_{\overline{s}}/H$, which is true. Now consider the factorization $[X/G] \to Y \to S$, where the first morphism is a gerbe. Since $[X/G] \to Y$ is flat and $[X/G] \to S$ is smooth, it follows that $Y \to S$ is smooth. Furthermore, the formation of $Y$ commutes with base change on $S$, and over every geometric point $\overline{s} \to S$ we have that the fiber $Y_{\overline{s}}$ is $\overline{s}$ (because the fiber $[X_{\overline{s}}/G_{\overline{s}}]$ is $BH$, which is a gerbe over $\overline{s}$). Hence we see that $Y \to S$ is a universally injective etale morphism, and hence it is an open immersion. Since $Y \to S$ is also surjective, we conclude that $Y=S$. QED
Proof of the Statement. At this point we know that $[X/G] \to S$ is a gerbe, and by the smoothness of $X$ it admits sections etale locally on $S$. After passing to an etale cover of $S$, we may fix a section of $X \to S$ and we obtain an isomorphism $[X/G] \cong BH$ for some flat subgroup scheme $H \subset G$ (the stabilizer of the section). At this point, to conclude that $X$ is a $G$-torsor we just need to show that $H \to S$ is the trivial group scheme. Since we are in characteristic $0$, the fibers of $H \to S$ are automatically regular, and hence $H \to S$ is smooth. Suppose for the sake of contradicton that $H \to S$ was nontrivial. Then there would be an etale $S$-scheme $T \to S$ and a $T$ point $h \in H(T)$ that is not the identity. Note that $h$ would fix the corresponding distinguished section $x \in X(T)$ induced by the trivial section $S \to BH$ under the isomorphism $[X/G] \cong BH$. Viewing $h \in G(T)$, this would contradict the freeness of the action of $G(T)$ on $X(T)$.QED