= Solution
Fix a prime $\mathfrak q$ of $E$ over $\mathfrak p$, and a prime $\mathfrak r$ of $LE$ over $\mathfrak q$. Let $\mathfrak P$ be its restriction to $L$. Inside the completed compositum, put
$$
F=K_{\mathfrak p},\qquad J=E_{\mathfrak q},\qquad U=L_{\mathfrak P}.
$$
The assumption means $U/F$ is an <unramified extension>. Its <residue degree> is some $f$; the completed extension $(LE)_{\mathfrak r}/J$ is the corresponding compositum $UJ/J$. It suffices to prove that this is <unramified>, regardless of the <ramification> of $J/F$.
Here is a proof of <unramified extensions remain unramified under base change> using the classification already established. Choose an integral generator $\alpha$ of $U/F$ whose monic <minimal polynomial> $g\in\mathcal O_F[X]$ has irreducible separable reduction of degree $f$. Write $k_F=\mathbb F_q$ and $k_J=\mathbb F_{q^r}$. Let $J_f/J$ be the degree-$f$ <unramified extension>. Its <residue field> $\mathbb F_{q^{rf}}$ contains $\mathbb F_{q^f}$, so $\bar g$ has a simple root there. <Hensel lemma> lifts this root to a root of $g$ in $J_f$. This gives an $F$-embedding $U\hookrightarrow J_f$.
An <unramified extension> of a <local field> is <Galois>, so all $F$-embeddings of $U$ into a fixed algebraic closure have the same image. The original $U$ therefore lies in $J_f$, and $UJ$ is an intermediate field of the <unramified extension> $J_f/J$. Its <ramification index> is one. Since $\mathfrak q$ and $\mathfrak r$ were arbitrary,
$$
\boxed{e(\mathfrak r/\mathfrak q)=1\text{ for every prime above }\mathfrak p.}
$$
This proof also explains why no linear-disjointness assumption on $L$ and $E$, and no assumption that $E/K$ is <unramified>, is needed.
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