Try consider $p=2^{16}+1$ is a prime, take one primitive root called $g$. Notice that every primitive root must congruent to $g^k\pmod{p}$ where $k$ is an odd integer.

The statement is only true if $p$ is not a Fermat prime, i.e. $p \neq 2^{2^n}+1$ for any positive integer $n$. Writing $a \equiv g^x, b \equiv g^y, c \equiv g^z \pmod{p}$ for some primitive root $g$ and integers $x,y,z$, we see that the statement is equivalent to the following: Do there exist integers $x, y, z$ such that $\gcd(x, p-1) = \gcd(y, p-1) = \gcd(z, p-1) = 1$ such that $\gcd(x+y+z, p-1) > 1$? And now, I claim that this is true if and only if $p$ is not a Fermat prime. First, assume that $p$ was not a Fermat prime. Then, there exists an odd prime $q$ dividing $p-1$. Choose $x\equiv 1\pmod{q}, y\equiv 1\pmod{q}, z\equiv -2\pmod{q}$, and $x, y, z \equiv 1\pmod{\frac{p-1}{q^{\nu_q(p-1)}}}$ (we can do this by CRT). Then, the gcd condition is upheld, but $x+y+z\equiv 0\pmod{q}$, so $q\mid \gcd(x+y+z, p-1)$, so $\gcd(x+y+z, p-1) > 1$. Thus, it is possible when $p$ is not a fermat prime. If $p$ is a fermat prime, then $p-1 = 2^{2^n}$, and so the gcd condition is equivalent to $x,y,z$ being odd. Then, $x+y+z$ is also odd, so we must necessarily have $\gcd(x+y+z, p-1) = 1$. So it is not possible when $p$ is a fermat prime. Thus, we've characterized all such $p$ for which this works, as desired. $\blacksquare$

Fun fact: If the constant $2023$ has been changed with another constant bigger than $65537$ then this problem would be equivalent to an open conjecture.