I'm trying to prove the following conjecture.
Conjecture. Let $p \equiv -1\!\pmod{6}$ be a prime, and let $a,b > p$ be integers with $p \nmid ab(a+b)$. Then $$ \sum_{r=0}^{p-3} \Biggl(\!\binom{p-1}{r+1}+(-1)^r\!\Biggr)\,a^rb^{p-3-r} \not\equiv 0 \pmod{p^2}. $$
Any hints would be greatly appreciated!
Note that the “coefficient” is divisible by $p$ in all terms of the summation.
After some tidying up, your conjecture is equivalent to $$(a+b)^p \not \equiv a^p + b^p \pmod{p^2}. $$
However, $4^{59} \equiv 3^{59} + 1 \pmod{59^2}$.