Let $\bar{E}$ be the norm completion of $E$, which is a Banach space. Then we can consider $E$ as a dense linear subspace of $\bar{E}$, where the subspace topology and the norm topology on $E$ coincide. In particular, since this topology is homeomorphic to $F$, it is completely metrizable, so $E$ is a $G_\delta$ in $\bar{E}$ (Kechris, Classical Descriptive Set Theory, Theorem 3.11). Being dense, by the Baire category theorem $E$ is nonmeager in $\bar{E}$. But by the Pettis lemma (Kechris Theorem 9.9), any subspace of the Banach space $\bar{E}$ which is nonmeager and has the Baire property (which a $G_\delta$ certainly does) must be $\bar{E}$ itself. Since $E$ is its own norm completion, its norm must be complete.

I think I did not use separability anywhere.

Let $\bar{E}$ be the norm completion of $E$, which is a Banach space. Then we can consider $E$ as a dense linear subspace of $\bar{E}$, where the subspace topology and the norm topology on $E$ coincide. In particular, since this topology is homeomorphic to $F$, it is completely metrizable, so $E$ is a $G_\delta$ in $\bar{E}$ (Kechris, Classical Descriptive Set Theory, Theorem 3.11). As a dense $G_\delta$, in particular $E$ is comeager in $\bar{E}$. If $x \in \bar{E} \setminus E$, then $E, E+x$ are disjoint comeager subsets of $\bar{E}$, which is absurd by the Baire category theorem. So $E = \bar{E}$ and thus the norm on $E$ is complete.

I think I did not use separability anywhere.

First suppose $E,F$ are separable. Let $\bar{E}$ be the norm completion of $E$, which is a separable Banach space. Then we can consider $E$ as a dense linear subspace of $\bar{E}$, where the subspace topology and the norm topology on $E$ coincide. In particular, since this topology is homeomorphic to $F$, it is completely metrizable, so $E$ is a $G_\delta$ in $\bar{E}$ (Kechris, Classical Descriptive Set Theory, Theorem 3.11). Being dense, by the Baire category theorem $E$ is nonmeager in $\bar{E}$. But by the Pettis lemma (Kechris Theorem 9.9), any subspace of the separable Banach space $\bar{E}$ which is nonmeager and has the Baire property (which a $G_\delta$ certainly does) must be $\bar{E}$ itself. Since $E$ is its own norm completion, its norm must be complete.

Still thinking about the non-separable case.

Let $\bar{E}$ be the norm completion of $E$, which is a Banach space. Then we can consider $E$ as a dense linear subspace of $\bar{E}$, where the subspace topology and the norm topology on $E$ coincide. In particular, since this topology is homeomorphic to $F$, it is completely metrizable, so $E$ is a $G_\delta$ in $\bar{E}$ (Kechris, Classical Descriptive Set Theory, Theorem 3.11). Being dense, by the Baire category theorem $E$ is nonmeager in $\bar{E}$. But by the Pettis lemma (Kechris Theorem 9.9), any subspace of the Banach space $\bar{E}$ which is nonmeager and has the Baire property (which a $G_\delta$ certainly does) must be $\bar{E}$ itself. Since $E$ is its own norm completion, its norm must be complete.

I think I did not use separability anywhere.