There is an approach that does not need any special properties of $g$ or $P$, that does not need any radicals and that does not need a big amount of manipulation of symmetric polynomials.
Let $A$ be a matrix with eigenvalues $x_1,\ldots, x_n$, where $n$ is the degree of $P$ (take the companion matrix of $P$). Then $g(x_1),\ldots,g(x_n)$ are the eigenvalues of $g(A)$. Let $Q$ be the characteristic polynomial of $g(A)$.
$$
Q(t):= \chi_{g(A)}(t) = t^n+q_{n-1}t^{n-1}+\ldots+q_0
$$
As $g(x_1),\ldots,g(x_n)$ are the roots of this polynomial, we have
\begin{eqnarray*}
e_1(g(x_1),\ldots,g(x_n)) & = \sum_i g(x_i) = & -q_{n-1} = Tr(g(A)) \\
e_2(g(x_1),\ldots,g(x_n)) & = & q_{n-2} \\
\vdots \\
e_{n-1} (g(x_1),\ldots,g(x_n)) & =& (-1)^{n-1} q_1 \\
e_n(g(x_1),\ldots,g(x_n)) & = \prod_i g(x_i) = & (-1)^n q_0 = \det(g(A))
\end{eqnarray*}
$(e_1,\ldots,e_n$ are the elementary symmetric polynomials.)