Call $a$ is good if there is such a prime $p$. I claim $a$ is good unless $a\in\{-2,1\}$. Assume $a$ is good (with some prime $p$), and denote the given sum by $S$. Note that $p\nmid a$: otherwise $p\mid (p-1)(p-2)/2$, which is not possible. Now, we observe, using binomial theorem \[ Sa^2 + 1+(p-1)a+(p-1)a^{p-2}+a^{p-1} = (1+a)^{p-1}\implies S=\frac{(1+a)^{p-1} - a^{p-1}-(p-1)a^{p-2}-(p-1)a-1}{a^2}. \]In particular, $p\mid S$ implies \[ p\mid (1+a)^{p-1} - a^{p-1}-(p-1)a^{p-2}-(p-1)a-1. \]Now, if $a\equiv -1\pmod{p}$ then we obtain $p=2$, not possible. Hence $a\not\equiv -1,0\pmod{p}$. Using this, Fermat's theorem implies $(1+a)^{p-1}\equiv a^{p-1}\pmod{p}$, hence $p\mid S$ if and only if \[ a^{p-2}+a+1\equiv 0\pmod{p}\iff p\mid a^2+a+1. \]Noting $a^2+a+1$ is always odd, such a $p$ always exists unless $a^2+a+1$ is a power of three. As $a^2+a+1$ is not divisible by $9$ (otherwise $9\mid (2a+1)^2+3$ and $-3$ is not a quadratic residue modulo $9$), $a$ is good unless $a^2+a+1=3$, that is unless $a\in\{-2,1\}$.