This is to complement the excellent answer by Alexandre Eremenko by showing that the only root of the equation $g(z)=1$ in $z$ with $|z|\le1$ is $1$.
Indeed, we have $g(z)=\sum_0^\infty a_nz^n$, where $a_n\ge0$ for all $n$, and $a_0a_1>0$. So, for any $z\ne1$ with $|z|\le1$ we have $|a_0+a_1z|<a_0+a_1$ and hence $$|g(z)|\le|a_0+a_1z|+\Big|\sum_2^\infty a_nz^n\Big|<a_0+a_1+\sum_2^\infty a_n=g(1)=1,$$ which implies the claim.