Solution:
Let us rewrite the given equation as follows:
$$a b + \sqrt{a^{2} + b} \sqrt{b^{2} + a} = -\sqrt{a b + 1}.$$
Squaring this gives us
$$\begin{aligned} a^{2} b^{2} + 2 a b \sqrt{a^{2} + b} \sqrt{b^{2} + a} + (a^{2} + b)(b^{2} + a) & = a b + 1 \\ (a^{2} b^{2} + a^{3}) + 2 a b \sqrt{a^{2} + b} \sqrt{b^{2} + a} + (a^{2} b^{2} + b^{3}) & = 1 \\ (a \sqrt{b^{2} + a} + b \sqrt{a^{2} + b})^{2} & = 1 \\ a \sqrt{b^{2} + a} + b \sqrt{a^{2} + b} & = \pm 1. \end{aligned}$$
Next, we show that $a \sqrt{b^{2} + a} + b \sqrt{a^{2} + b} > 0$. Note that
$$a b = -\sqrt{a b + 1} - \sqrt{a^{2} + b} \cdot \sqrt{b^{2} + a} < 0$$
so $a$ and $b$ have opposite signs. Without loss of generality, we may assume $a > 0 > b$. Then rewrite
$$a \sqrt{b^{2} + a} + b \sqrt{a^{2} + b} = a (\sqrt{b^{2} + a} + b) - b (a - \sqrt{a^{2} + b})$$
and, since $\sqrt{b^{2} + a} + b$ and $a - \sqrt{a^{2} + b}$ are both positive, the expression above is positive. Therefore,
$$a \sqrt{b^{2} + a} + b \sqrt{a^{2} + b} = 1,$$
and the proof is finished.