Yes. The first thing to notice is that the Zeta function is the Artin L-function associated to the trivial representation of $Gal(E/K)$ i.e.
$\zeta_K(s) = L(s, \mathbb{C}, E/K),$
where $\mathbb{C}$ is endowed with the trivial action of $\mathrm{Gal}(E/K).$
Let $G = \mathrm{Gal}(E/F)$ and $H = \mathrm{Gal}(E/K).$ Now recall that the $L$-function attached to any representation of $\rho$ of $H$ is equal to the $L$-function associated to the induced representation $\rho^G_H$ of $G.$
In particular,
$L(s, \mathbb{C}, E/K) = L(s, \mathbb{C}[G]\otimes_{\mathbb{C}[H]}\mathbb{C}, E/F).$
Therefore, if $\chi_1,...,\chi_n$ are the irreducible characters of $G$ and $\chi$ is the character of $\mathbb{C}[G]\otimes_{\mathbb{C}[H]}\mathbb{C},$
$L(s, \mathbb{C}[G]\otimes_{\mathbb{C}[H]}\mathbb{C}, E/F) =L(s, \chi,E/F) = \displaystyle\prod_{i=1}^nL(s, \chi_i,E/F)^{\langle \chi_i, \chi \rangle},$
And so,
$\zeta_K(s) = \displaystyle\prod_{i=1}^nL(s, \chi_i,E/F)^{\langle \chi_i, \chi \rangle}.$
In the case $H$ and $G$ are explicit, the characters $\chi_1,...,\chi_n$ and $\chi$ are easily calculated and thus give an explicit factorization of $\zeta_K(s).$