Assume the contrary, then we have that $p \nmid a$ and $p\nmid b$ (as $p$ dividing one of them would mean that $p$ divides the other), and so we have that $$\frac{a^2 + ab + b^2}{b^2}\equiv \left(\frac{a}{b}\right)^2 + \left(\frac{a}{b}\right) + 1 \equiv 0\pmod{p}.$$This means that there exists an $r\in \mathbb{F}_{p}$ such that $r^2 + r + 1 \equiv 0\pmod{p}$, but since the roots of $r^2 + r + 1$ are $\frac{-1\pm \sqrt{-3}}{2}$, we must have $-3$ a quadratic residue modulo $p$. However, since $p \equiv 2\pmod{3}$, we have by Quadratic Reciprocity and the fact that $\left(\frac{-1}{p}\right) = (-1)^{\frac{p-1}{2}}$ that $$\left(\frac{-3}{p}\right) = \left(\frac{-1}{p}\right)\left(\frac{3}{p}\right) = (-1)^{\frac{p-1}{2}}\cdot (-1)^{\frac{3-1}{2}\cdot \frac{p-1}{2}} \left(\frac{p}{3}\right) = (-1)^{p-1}\left(\frac{p}{3}\right) = \left(\frac{2}{3}\right) = -1,$$a contradiction. (Here, $\left(\frac{a}{p}\right)$ denotes the Legendre Symbol)

@above what if p|a-b, you didn't account for that case.