After Nik Weaver answered the question and Bill Johnson pointed out in a comment that what one needs is actually a part of the usual proof of the open mapping theorem, I thought about it once again, and I think it is worthwhile to explicitly remark that the following more general assertion is true:

Theorem. Let $E$, $F$ be Banach space and suppose that $E$ embeds densely and continuously into $F$ (so we may consider $E$ as a subspace of $F$). Suppose that for each $f \in F$ there exists a sequence $(e_n)$ in $E$ which converges to $f$ with respect to $\|\cdot\|_F$ and which is bounded with respect to $\|\cdot\|_E$. Then $E = F$.

Discussion. The point here is that, in constrast to the statement in the original question, we do not assume a priori that the bound $\sup_n \|e_n\|_E$ is uniform with respect to $\|f\|_F$.

Proof. Actually, this is almost exactly the proof of the open mapping theorem, but with a (very) small additional perturbation argument. The details are as follows:

Let $j$ denote the embedding of $E$ into $F$. Let $B_E(e,r)$ denote the open ball in $E$ with radius $r$ and center $e$, and likewise for $F$. It follows from the assumption that $$ \bigcup_{n \in \mathbb{N}} \overline{ j B_E(0,n)}^F = F. $$ Due to Baire's theorem we can thus find an integer $n \in \mathbb{N}$ such that $\overline{ j B_E(0,n)}^F$ has non-empty interior in $F$, i.e. it contains a ball $B_F(f,\varepsilon)$.

Since $j(E)$ is dense in $F$, we can choose $e \in E$ such that $j(e)$ is closer than $\varepsilon / 2$ to $f$ (with respect to $\|\cdot\|_F$). (This is the "perturbation argument" mentioned above - well, "argument" is quite an exaggeration...) Hence, $$ \overline{ j B_E(0,n)}^F \supseteq B_F(f,\varepsilon) \supseteq B_F(j(e), \varepsilon/2). $$ Now, we proceed again as in the proof of the open mapping theorem: We can write each $x \in B_F(0, \varepsilon/2)$ as $x = (x+j(e)) - j(e)$, and the vector in brackets is in $B_F(j(e), \varepsilon/2)$ and can thus be approximated (wrt $\|\cdot\|_F$) by a sequence $(j(e_k))$ with $\|e_k\|_E < n$. Since $j(e_k - e)$ approximates $x$ (wrt $\|\cdot\|_F$), we obtain $x \in \overline{ j B_E(0,n + \|e\|)}^F$.

We proved that the $F$-closure of $j B_E(0,n + \|e\|)$ contains $B_F(0, \varepsilon/2)$, so it follows from the small open mapping theorem (see Bill Johnson's comment to Nik Weaver's answer) that $j B_E(0,n + \|e\|)$ itself contains an open ball in $F$ centred at $0$. Hence, $j(E) = F$.

After Nik Weaver answered the question and Bill Johnson pointed out in a comment that what one needs is actually a part of the usual proof of the open mapping theorem, I thought about it once again, and I think it is worthwhile to explicitly remark that the following more general assertion is true:

Theorem. Let $E$, $F$ be Banach space and suppose that $E$ embeds densely and continuously into $F$ (so we may consider $E$ as a subspace of $F$). Suppose that for each $f \in F$ there exists a sequence $(e_n)$ in $E$ which converges to $f$ with respect to $\|\cdot\|_F$ and which is bounded with respect to $\|\cdot\|_E$. Then $E = F$.

Discussion. The point here is that, in constrast to the statement in the original question, we do not assume a priori that the bound $\sup_n \|e_n\|_E$ is uniform with respect to $\|f\|_F$.

Proof. Actually, this is almost exactly the proof of the open mapping theorem, but with a (very) small additional perturbation argument. The details are as follows:

Let $j$ denote the embedding of $E$ into $F$. Let $B_E(e,r)$ denote the open ball in $E$ with radius $r$ and center $e$, and likewise for $F$. It follows from the assumption that $$ \bigcup_{n \in \mathbb{N}} \overline{ j B_E(0,n)}^F = F. $$ Due to Baire's theorem we can thus find an integer $n \in \mathbb{N}$ such that $\overline{ j B_E(0,n)}^F$ has non-empty interior in $F$, i.e. it contains a ball $B_F(f,\varepsilon)$.

Since $j(E)$ is dense in $F$, we can choose $e \in E$ such that $j(e)$ is closer than $\varepsilon / 2$ to $f$ (with respect to $\|\cdot\|_F$). (This is the "perturbation argument" mentioned above - well, "argument" is quite an exaggeration...) Hence, $$ \overline{ j B_E(0,n)}^F \supseteq B_F(f,\varepsilon) \supseteq B_F(j(e), \varepsilon/2). $$ Now, we proceed again as in the proof of the open mapping theorem: We can write each $x \in B_F(0, \varepsilon/2)$ as $x = (x+j(e)) - j(e)$, and the vector in brackets is in $B_F(j(e), \varepsilon/2)$ and can thus be approximated (wrt $\|\cdot\|_F$) by a sequence $(j(e_k))$ with $\|e_k\|_E < n$. Since $j(e_k - e)$ approximates $x$ (wrt $\|\cdot\|_F$), we obtain $x \in \overline{ j B_E(0,n + \|e\|)}^F$.

We proved that the $F$-closure of $j B_E(0,n + \|e\|)$ contains $B_F(0, \varepsilon/2)$, so it follows from the small open mapping theorem (see Bill Johnson's comment to Nik Weaver's answer) that $j B_E(0,n + \|e\|)$ itself contains an open ball in $F$ centred at $0$. Hence, $j(E) = F$.

Edit 2023-05-02. I found a reference where the theorem above along with a different proof is explicitly stated: Theorem 5 in "Evgeniĭ A. Lifshits: Ideally convex sets (1970)" (link to zbMATH)