Let $\varphi\colon S\to S'$ be an isometry. Since $\mathbb{H}$ is simply connected, $\varphi$ lifts to a map $\widetilde{\varphi}\colon \mathbb{H}\to \mathbb{H}$ making the following diagram commute: $ \begin{array}{ccc} \mathbb{H} & \xrightarrow{\widetilde{\varphi}} & \mathbb{H} \\ \downarrow & & \downarrow \\ S & \xrightarrow{\varphi} & S' \end{array} $ Then $\widetilde{\varphi}$ is a local isometry. Since $\mathbb{H}$ is simply connected and geodesically complete, it follows that $\widetilde{\varphi}$ is an isometry. We claim that $\widetilde{\varphi}^{-1}\;\Gamma'\,\widetilde{\varphi} = \Gamma$.
Let $\gamma'\in\Gamma'$, and let $p\colon\mathbb{H}\to S$ and $p'\colon\mathbb{H}\to S'$ be the covering maps. We know that $\gamma'$ is a covering transformation for $p'$, i.e. $p'\gamma'=p'$. Since $p'\widetilde{\varphi} = \varphi p$, we have $ p\widetilde{\varphi}^{-1}\gamma'\widetilde{\varphi} \,=\, \varphi^{-1}p' \gamma'\widetilde{\varphi} \,=\, \varphi^{-1} p' \widetilde{\varphi} = p. $ Thus $\widetilde{\varphi}^{-1}\gamma'\widetilde{\varphi}$ is a covering transformation for $p$, so $\widetilde{\varphi}^{-1}\gamma'\widetilde{\varphi} \in \Gamma$. This proves that $\widetilde{\varphi}^{-1}\Gamma'\widetilde{\varphi} \leq \Gamma$, and a similar argument shows that $\Gamma \leq \widetilde{\varphi}^{-1}\Gamma'\widetilde{\varphi}$.