Comparison of Topologies via Closed Sets
No, but you can conclude that $\tau_e\supset\tau$. In fact\begin{align}A\in\tau&\iff A^\complement\text{ is $\tau$-closed}\\&\implies A^\complement\text{ is $\tau_e$-closed}\\&\iff A\in\tau_e.\end{align}
No, but you can conclude that $\tau_e\supset\tau$. In fact\begin{align}A\in\tau&\iff A^\complement\text{ is $\tau$-closed}\\&\implies A^\complement\text{ is $\tau_e$-closed}\\&\iff A\in\tau_e.\end{align}