You need to assume that $K$ is Galois over $K\cap E$; if that is the case, then the isomorphism drops out of the Galois correspondence and the Isomorphism Theorems.
We can replace $F$ with $K\cap E$, so that we are in the following situation:
- $KE/F$ is finite Galois;
- $K\cap E = F$.
If $G=\mathrm{Gal}(KE/F)$, let $M$ be the subgroup corresponding to $K$ and $N$ the subgroup corresponding to $E$. Then $M\cap N=\{e\}$ (since $KE$ is the field "on top"), and $\langle M,N\rangle = G$.
Now, $K$ is Galois over $F$ if and only if $M$ is normal in $\langle M,N\rangle$.
In particular, if $K$ is Galois over $F$, then $M\triangleleft \langle M,N\rangle$, so $\langle M,N\rangle = MN$. Thus, $N\cap M$ is normal in $N$, and by the isomorphism theorems we have that $\frac{G}{M} =\frac{MN}{M} \cong \frac{N}{N\cap M} \cong N.$ Now, $\frac{G}{M}\cong\mathrm{Gal}(K/F)$; and $N=\mathrm{Gal}(KE/E)$; so we get the isomorphism if $K$ is Galois.
For an example showing that the given conditions do not imply that $K$ is Galois over $K\cap E$, let $F=\mathbb{Q}$, $K=\mathbb{Q}[\sqrt[3]{2}]$, and $E=\mathbb{Q}(\zeta)$, where $\zeta$ is a primitive cubic root of unity; then $K\cap E=\mathbb{Q}$. Then $KE$ is the splitting field of $x^3-2$ over $\mathbb{Q}$, hence is Galois. Even though $KE$ is Galois over $E$, $K$ is not Galois over $\mathbb{Q}$, so you cannot have the claimed isomorphism.