Since this is near the top again, I'll add what seems natural to me. Let $F/\mathbb{Q}$ be an $S_3$ extension, and let $C$ and $K$ be the cubic and quadratic subfields. Then the $\zeta$ functions of these number fields have Euler products.
Let $p \in \mathbb{Z}$ be a prime which is not ramified in $F$. Then the corresponding Euler factor in each of the fields $\mathbb{Q}$, $K$, $C$ and $F$ is dictated by whether the Frobenius element has conjugacy class $e$, $(12)$ or $(123)$. Here is the formula for the Euler factor in each case:
$$\begin{array}{|c|c|c|c|} \hline & e & (12) & (123) \\ \hline \zeta_\mathbb{Q} & (1-p^{-s})^{-1} & (1-p^{-s})^{-1} & (1-p^{-s})^{-1} \\ \hline \zeta_K & (1-p^{-s})^{-2} & (1-p^{-s})^{-1}(1+p^{-s})^{-1}&(1-p^{-s})^{-2} \\ \hline \zeta_C & (1-p^{-s})^{-3} & (1-p^{-s})^{-2} (1+p^{-s})^{-1}&(1-p^{-s})^{-1}(1+p^{-s}+p^{-2s})^{-1} \\ \hline \zeta_F & (1-p^{-s})^{-6} &(1-p^{-s})^{-3}(1+p^{-s})^{-3}& (1-p^{-s})^{-2}(1+p^{-s}+p^{-2s})^{-2}\\ \hline \end{array}$$
Looking at this table makes it natural to imagine additional rows for $L$-functions $L_1$ and $L_2$ with $\zeta_K = \zeta_{\mathbb{Q}} L_1$, $\zeta_C = \zeta_{\mathbb{Q}} L_2$ and $\zeta_F = \zeta_{\mathbb{Q}} L_1 L_2^2$ and Euler factors $$\begin{array}{|c|c|c|c|} \hline & e & (12) & (123) \\ \hline L_1 & (1-p^{-s})^{-1} &(1+p^{-s})^{-1} &(1-p^{-s})^{-1} \\ \hline L_2 &(1-p^{-s})^{-2} &(1-p^{-s})^{-1} (1+p^{-s})^{-1}& (1+p^{-s}+p^{-2s})^{-1}\\ \hline \end{array}$$ (It especially helps that Hecke and Dirichlet already studied $L_1$.)
We now have three things associated to $S_3$, namely $\zeta_{\mathbb{Q}}$, $L_1$ and $L_2$, and more complicated things are made from them. In particular, $\zeta_F$ has 1 copy of the "trivial" $\zeta_{\mathbb{Q}}$, 1 copy of $L_1$ and $2$ copies of $L_2$. That already screams representation theory to me. Then when I notice that the degree $d$ representations gives degree $d$ polynomials in $p^{-s}$ in the denominator, what could be more natural than a characteristic polynomial?