Eliminating $c$ gives

$$a+b+c-a b c=a+b+(1-a b) c=a+b+\frac{(1-a b)(3-a b)}{a+b}$$

Put $x=\sqrt{a b}$. Then $a+b \geq 2 x$, and since $1<x^{2}<3, \frac{(1-a b)(3-a b)}{a+b} \geq \frac{\left(1-x^{2}\right)\left(3-x^{2}\right)}{2 x}$.

It then suffices to prove that

$$2 x+\frac{\left(1-x^{2}\right)\left(3-x^{2}\right)}{2 x} \geq 2$$

This last inequality follows from the arithmetic-geometric means inequality

$$2 x+\frac{\left(1-x^{2}\right)\left(3-x^{2}\right)}{2 x}=\frac{3+x^{4}}{2 x}=\frac{1}{2 x}+\frac{1}{2 x}+\frac{1}{2 x}+\frac{x^{3}}{2} \geq 4\left(\frac{1}{16}\right)^{\frac{1}{4}}=2$$