Eigenvalue of Cayley graph as character sum

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I'm working through Audrey Terras' Fourier Analysis on Finite Groups with Applications, and at the end of the chapter on induced representations, there is an exercise which reads:

Let $$S = \bigcup_{i=1}^s C_i$$ where $C_i$ is a conjugacy class in a finite group $G$. Suppose $S$ is symmetric; that is, $x\in S$ implies $x^{-1}\in S$. Consider the Cayley graph $X(G,S)$. Show that the eigenvalues of the adjacency matrix of this graph have the form $$\lambda_\pi = \frac{1}{d_\pi}\sum_{s\in S}\chi_\pi(s),$$ where $\pi \in \widehat G$, and $d_\pi=$ the degree of $\pi$.

I tried to mimic the proof for the abelian case, which is straightforward as you can use the fact that $\chi_\pi$ is always a homomorphism in that case. So I'd like to know if there is a way to compute it directly for the nonabelian case, taking $A$ to be the adjacency matrix of the Cayley graph, and $v_\pi=(\chi_\pi(g_1),\cdots,\chi_\pi(g_n))$ for $G=\{g_i\}_{i=1}^n$. Then where I get stuck in the computation is

$$ \begin{split} (Av_\pi)_x & = \sum_{y\in G}A_{xy}(v_\pi)_y \\ & = \sum_{x^{-1}y\in S}\chi_\pi(y) \\ & = \sum_{s\in S} \chi_\pi(xs) \\ & = \sum_{s\in S} \sum_{i=1}^{d_\pi} \pi_{ii}(xs) \end{split} $$

I did try this for the two-dimensional character of $S_3$, with $S$={(123),(132)} to try and get some intuition as to why this is true and maybe a way to prove it, but that didn't seem to provide any insight.

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Hints: Define $$ a=\sum_{s\in S}s\in\mathbb C[G] $$ and let $L_a:\mathbb C[G]\to\mathbb C[G]$ denote left multiplication by $a$. Then the matrix of $L_a$ with respect to the basis $G$ of $\mathbb C[G]$ is $A$. Now use the isomorphism $$ \mathbb C[G]\cong\bigoplus_V\mathrm{End}_{\mathbb C}(V) $$ where the sum is over all irreducible representations.