Disclaimer: When I came up with this question yesterday, I suspected it to be trivial (trivially true or trivially false). Then it kept me awake several hours tonight... (I still hope, though, this is just due to my ignorance.)

Question. Let $E,F$ be Banach space and suppose that $E$ embeds densely and continuously into $F$ (so we consider $E$ as a subspace of $F$ from now on). Assume that there exists a constant $M \in (0,\infty)$ with the following property:

For each $f \in F$ we can find a sequence $(e_n)$ in $E$ that converges to $f$ with respect to $\|\cdot\|_F$ and that satisfies $\|e_n\|_E \le M \|f\|_F$.

Does it follow that $E = F$?

Remark. I first thought the answer should be yes due to some application of the open mapping theorem: clearly, it suffices to show that $\|\cdot\|_E$ and $\|\cdot\|_F$ are equivalent on $E$, and by the open mapping theorem this is true iff $\|\cdot\|_F$ is complete on $E$; but I wasn't able to prove that latter property.

Am I overlooking some simple argument, or a simple counterexample?

Edit. It is probably worthwhile to note the following fact:

As observed by Nate Eldredge in a (now deleted) comment, it is easy to see that the answer is "yes" if $E$ is reflexive: Given $f \in F$ and $(e_n) \subseteq E$ as above, we can choose a subsequence of $(e_n)$ that converges weakly (in $E$) to a vector $e \in E$. For each $f' \in F'$ this implies that $\langle f', f\rangle = \langle f', e\rangle$, so $f = e \in E$.