Let $G\to S$ be a group scheme acting on a scheme $X\to S$. Define a prestack $[X/G]$ whose objects over an $S$-scheme $T$ are pairs $(P\to T, P\xrightarrow{\varphi} X)$ where $P\to T$ is a $G$-torsor and $\varphi$ is $G$-equivariant. Morphisms $(P\to T, P\xrightarrow{\varphi}X)\to(P'\to T', P'\xrightarrow{\varphi'}X)$ over a morphism $g:T'\to T$ are morphisms $P'\xrightarrow{f}P$ pulling back $g$ via $P\to T$.
I've seen several sources claim that the morphism $X\to[X/G]$ associated to the object $(G\times X\xrightarrow{p_2}X,G\times X\xrightarrow{\sigma}X)$ (via $2$-Yoneda) is a $G$-torsor. Since no topology has been mentioned so far, I believe this means that the map $G\times_SX\to X\times_{[X/G]}X$ is an isomorphism. Since I'm trying to prove that this is an isomorphism of prestacks (also called categories fibred in groupoids), it's enough to show that this induces an equivalence of categories over all fibres.
Given an $S$-scheme $T$, an object $T\to X$ maps to $(G\times_ST\to T, G\times_ST\to X)$ where the latter factors though $G\times_ST\to G\times_SX$ via the natural map $X\to[X/G]$. So the fibred product $(X\times_{[X/G]}X)(T)$ should consists of triples $(T\xrightarrow{\alpha}X,T\xrightarrow{\beta}X,\eta)$ where $\eta:(G\times_ST)_\alpha\to(G\times_ST)_\beta$ where $(G\times_ST)_\alpha = (G\times_SX)\times_{p_2,X,\alpha}T$ and similarly for $\beta$. How can I canonically associate to this a map $T\to G\times_SX$?
This seems sort of like the prestack associated to $G\times_SX$, but I don't even see how to formally construct a map $X\times_{[X/G]}X\to G\times_SX$, let alone write an inverse. Any help would be much appreciated.