Let $G = \mathbb{Z}/p \times \mathbb{Z}/p$ for $p$ odd. Then $H^3(BG; \mathbb{Z}) \cong \mathbb{Z}/p$ is not in the subring generated by Euler classes, since the non-trivial irreducible representations are of rank $2$.

Let $G = \mathbb{Z}/p \times \mathbb{Z}/p$ for $p$ odd. Then $H^3(BG; \mathbb{Z}) \cong \mathbb{Z}/p$ is not in the subring generated by Euler classes, since the non-trivial irreducible representations are of rank $2$.

EDIT: The problem of determining the subring of $H^*(BG; \mathbb{Z})$ generated by all Chern classes has been much studied. You might start with Atiyah's 1961 paper.