The result is true, regardless of characteristic.

Lemma: Let A be a k algebra and M a semi-simple A-module which is finite dimensional as a k-algebra. Then the image of A in $\mathrm{End}_k(M)$ is a semi-simple ring.

So, by the Artin-Wedderburn theorem, this image is a direct sum of matrix algebras over division rings.

Proof: Call this image S. Since S is finite dimensional, it is artinian. Let $M= \bigoplus V_i$ and let $t=(t_i)$ lie in the Jacobson radical of S. For each $V_i$, the condition that $V_i$ is a simple A-module implies that it is a simple S-module. So t must act trivially on $V_i$, and thus $t_i=0$. But we have proved this for all i, so $t=0$ and the Jacobson radical of S is trivial. QED.

We can now reduce to the case that A=S, and is a direct sum of division rings. Say $A = \bigoplus \mathrm{Mat}_{n_i}(\Delta_i)$. So every A-module is of the form
$$\bigoplus (\Delta_i^{n_i})^{k_i}$$
for some $k_i$ and the corresponding characteristic polynomial is
$$\chi(\lambda, g) = \prod \chi_i(\lambda, g_i)^{k_i}$$
where, for $h \in \mathrm{Mat}_{n_i}(\Delta_i)$, the polynomial $\chi_i(\lambda, h)$ is the characteristic polynomial of $h$ acting on $\Delta_i^{n_i}$.

A much better argument, suggested by buzzard's comment below. (I am not sure whether or not the original can be fixed.) Let $M = \bigoplus (\Delta_i^{n_i})^{k_i}$ and $N = (\Delta_i^{n_i})^{\ell_i}$. Suppose M is not a summand of N, so $k_i > \ell_i$ for some $i$. Let g be 1 on the i component and 0 everywhere else. Then the characteristic polynomials of M and N are of the form $(\lambda-1)^{n_i k_i} \lambda^{\bullet}$ and $(\lambda-1)^{n_i \ell_i} \lambda^{\bullet}$. So the former does not divide the latter.

~~We just need to show that the polynomials $\chi_i$, as polynomial functions on $\overline{k} \times A$, are relatively prime to each other. This is easy enough. Let $t_a$ be a basis for the $k$-linear functions on $\mathrm{Mat}_{n_i}(\Delta_i)$ and $u_b$ a basis for the $k$-linear functions on $\mathrm{Mat}_{n_j}(\Delta_j)$. Then $\chi_i(\lambda, g_i)$ is a homogenous polynomial in $\lambda$ and $t_a$; while $\chi_j$ in homogenous in $\lambda$ and $u_b$. Their GCD must be homogenous in both ways, hence, it is of the form $\lambda^m$.~~

~~
~~~~But, if $\lambda | \chi_i(\lambda)$, this means that there is no $g$ which acts invertibly on $\Delta_i^{n_i}$; contradicting that the identity does so act. So the GCD is 1, and the result is true.~~