The resulting topology is $\tau$, if the action of $G$ on $X$ is continuous (i.e. the multiplication map $X\times G\to X$ is continuous):

Let $\tilde X$ denote $(X\times G)/\sim$, and let $\phi:X\to \tilde X:x\mapsto[x,e]$ be the identification you mentioned (with the factors $X,G$ reversed for convenience). Then $\phi$ is clearly continuous. Let $\psi:\tilde X\to X$ be the inverse of $\phi$, i.e. $\psi([x,g]) = xg$. For an open subset $U$ of $X$, $\psi^{-1}(U) = \{(x,g):xg\in U, g\in G\}$. Pulling this back to $X\times G$ via the projection $X\times G\to\tilde X$ gives exactly the preimage of $U$ under the multiplication map $X\times G\to G$, which is open, and so $\psi^{-1}(U)$ is open in $\tilde X$. So $\phi$ is a homeomorphism.

If the action of $G$ on $X$ is continuous (i.e. the multiplication map $X\times G\to X$ is continuous) then the resulting topology is $\tau$:

Let $\tilde X$ denote $(X\times G)/\sim$, and let $\phi:X\to \tilde X:x\mapsto[x,e]$ be the identification you mentioned (with the factors $X,G$ reversed for convenience). Then $\phi$ is clearly continuous. Let $\psi:\tilde X\to X$ be the inverse of $\phi$, i.e. $\psi([x,g]) = xg$. For an open subset $U$ of $X$, $\psi^{-1}(U) = \{(x,g):xg\in U\}$. Pulling this back to $X\times G$ via the projection $X\times G\to\tilde X$ gives exactly the preimage of $U$ under the multiplication map $X\times G\to G$, which is open, and so $\psi^{-1}(U)$ is open in $\tilde X$. So $\phi$ is a homeomorphism.

The resulting topology is $\tau$, if the action of $G$ on $X$ is continuous (i.e. the multiplication map $X\times G\to X$ is continuous):

Let $\tilde X$ denote $(X\times G)/\sim$, and let $\phi:X\to \tilde X:x\mapsto[x,e]$ be the identification you mentioned (with the factors reversed for convenience). Then $\phi$ is clearly continuous. Let $\psi:\tilde X\to X$ be the inverse of $\phi$, i.e. $\psi([x,g]) = xg$. For an open subset $U$ of $X$, $\psi^{-1}(U) = \{(x,g):xg\in U, g\in G\}$. Pulling this back to $X\times G$ via the projection $X\times G\to\tilde X$ gives exactly the preimage of $U$ under the multiplication map $X\times G\to G$, which is open, and so $\psi^{-1}(U)$ is open in $\tilde X$. So $\phi$ is a homeomorphism.

The resulting topology is $\tau$, if the action of $G$ on $X$ is continuous (i.e. the multiplication map $X\times G\to X$ is continuous):

Let $\tilde X$ denote $(X\times G)/\sim$, and let $\phi:X\to \tilde X:x\mapsto[x,e]$ be the identification you mentioned (with the factors $X,G$ reversed for convenience). Then $\phi$ is clearly continuous. Let $\psi:\tilde X\to X$ be the inverse of $\phi$, i.e. $\psi([x,g]) = xg$. For an open subset $U$ of $X$, $\psi^{-1}(U) = \{(x,g):xg\in U, g\in G\}$. Pulling this back to $X\times G$ via the projection $X\times G\to\tilde X$ gives exactly the preimage of $U$ under the multiplication map $X\times G\to G$, which is open, and so $\psi^{-1}(U)$ is open in $\tilde X$. So $\phi$ is a homeomorphism.