For any abelian group $G$ your $\zeta(G)$ is equal to $1$. For the cyclic group of order $4$ your $P_G(z)$ is $z+z^2+z^4$. The zeros of this polynomial are zero, together with roughly $-0.68233$ and $0.34116\pm 1.16154i$. The last two of these are outside the unit disc.

Having said this, given that you have not checked the cyclic group of order four, I strongly suspect that this question is generated by AI, and would like to persuade you that this is not an appropriate way to proceed on this website.

For any abelian group $G$ your $\zeta(G)$ is equal to $1$. For the cyclic group of order $4$ your $P_G(z)$ is $z+z^2+z^4$. The zeros of this polynomial are zero, together with roughly $-0.68233$ and $0.34116\pm 1.16154i$. The last two of these are outside the unit disc.

For the smallest non-abelian simple group, $A_5$, the polynomial is $$\textstyle\frac{61}{15}z+\frac{25}{2}z^5+\frac{54}{5}z^6+\frac{50}{3}z^{10}+6z^{12}+5z^{15}+10z^{20}+15z^{30}+z^{60}$$ which also has zeros outside the unit disc, for example roughly $-1.09491+0.10363i$.

Having said this, given that you have not checked the smallest non-abelian simple group, I strongly suspect that this question is generated by AI, and would like to persuade you that this is not an appropriate way to proceed on this website.