Of course the correct answer is provided by Chris Wuthrich in the comments. But, let me comment a fun, ‘high-brow’ way of seeing this.
Choose an integer $p\ne N\geqslant 3$ and let $\mathcal{Y}(N)$ be the modular curve of level $\Gamma(N)$ over $\mathscr{O}_K$. In other words, it represents the following functor on $\mathscr{O}_K$-algebras $R$:
$$\mathcal{Y}(N)(R)=\left\{(\mathcal{E},\alpha):\begin{aligned}(1) & \quad \mathcal{E}\to\mathrm{Spec}(R)\text{ is an elliptic curve},\\ (2) & \quad \alpha\colon \underline{(\mathbb{Z}/N)}^2\xrightarrow{\sim}\mathcal{E}[N].\end{aligned}\right\}$$
There is a $j$-invariant map $j\colon \mathcal{Y}(N)\to \mathbf{A}^1_{\mathscr{O}_K}$ just given by $j(\mathcal{E},\alpha)=j(\mathcal{E})$.
Exercise: The map $j$ is finite.
Because $j$ is finite (and so proper) it follows that (just think about the valuative criterion!)
$$\widehat{\mathcal{Y}}(N)_\eta=j^{-1}((\hat{\mathbf{A}}^1_{\mathscr{O}_K})_\eta)=j^{-1}(\{|x|\leqslant 1\}),$$
where the completions are the $p$-adic ones.
Conclusion: We have the equality $\widehat{\mathcal{Y}}(N)_\eta=j^{-1}(\{|x|\leqslant 1\}).$
But $E/K$ is completable to a point $(E,\alpha)$ of $\widehat{\mathcal{Y}}(N)_\eta$ if and only if $E$ has potentially good reduction.
If potentially good reduction: for any model $\mathcal{E}$ over a finite extension, $\mathcal{E}[N]$ is étale (as $p\nmid N$), so after extending further we can complete to some $(\mathcal{E},\alpha)$.
If completable: if $E$ can be completed to an $L$-point of $\widehat{\mathcal{Y}}(N)_\eta(K)$ then by definition this extends to a $\mathcal{O}_L$-point $(\mathcal{E},\alpha)$, and $\mathcal{E}_L=E_L$.
(NB: if one is willing to use stacks, then you can avoid this extraneous level structure, but then you have to think about stacks and rigid spaces