Let $p = ab$ and $s = a + b$. We want to show that $p^2(s^2 - 2p - 2) \ge s(p-1)$, or equivalently, $p^2s^2 - (p-1)s - 2p^2(p+1) \ge 0$ when $s^2 \ge 4p > 0$.
If $0 < p \le 1$, then $p^2s - (p-1) \ge p^2 \cdot 2\sqrt{p} + (1-p) > 0$, and if $p \ge 1$, then $p^2s - (p-1) \ge (p^2 \cdot 2\sqrt{p} - p) + 1 > 0$. Therefore,
$$\begin{align*} p^2 s^2 - (p-1)s - 2p^2(p+1) &= s(p^2s - (p-1)) - 2p^2(p+1) \\ &\ge 2\sqrt{p}(p^2 \cdot 2\sqrt{p} - p + 1) - 2p^2(p+1) \\ &= 2p^3 - 2p\sqrt{p} - 2p^2 + 2\sqrt{p} \\ &= 2\sqrt{p}(p-1)(p\sqrt{p} - 1) \\ &\ge 0. \end{align*}$$