 Solution

Because the metric coefficients are independent of $t$ and $y$,are Killing vector fields. The mapssatisfy ${\varphi }_s\circ {\varphi }_r={\varphi }_{s+r}$. Moreover, $dx$ is unchanged and both $e^{x}dy$ and $dt+\sqrt{2}{e}^{x}dy$ are invariant, so ${\varphi }_s^{\ast }g=g$. Differentiating at $s=0$ produces the third Killing fieldGiven two points, first use ${\varphi }_s$ to match their $x$ coordinates, then translations generated by ${\partial }_y$ and ${\partial }_t$ to match $y$ and $t$. The isometry group therefore acts transitively, so the spacetime is a homogeneous space.