Solution:

The minimum is $3$, which is obtained for $(a, b, c) = (1, 1, -3)$ and permutations of this triple.

As $a b c$ is negative, the triple $(a, b, c)$ has either exactly one negative number or three negative numbers. Also, since $|a b c| \geq 3$, at least one of the three numbers has absolute value greater than $1$.

If all of $a, b, c$ were negative, the previous statement would contradict $a+b+c = -1$, hence exactly one of $a, b, c$ is negative.

WLOG let $c$ be the unique negative number. So $a, b > 0 > c$, as the value $0$ isn't possible by $|a b c| \geq 3$. Let $S$ be the given sum of fractions. We then have
$$\begin{aligned} S+3 & = \sum_{cyc} \frac{a b+1+a+b}{a+b} = \sum_{cyc} \frac{(a+1)(b+1)}{a+b} = \sum_{cyc} -\frac{(a+1)(b+1)}{c+1} \\ & \geq \sum_{cyc} |a+1| = (a+1)+(b+1)-(c+1) = 2a+2b+2 \end{aligned}$$
using AM-GM on the three pairs of summands respectively for the inequality. We can do this since $a+1, b+1 > 0$ and $-(c+1) = a+b > 0$, so every summand is positive.

So all we want to do now is show $a+b \geq 2$, to conclude $S \geq 3$. From the two given conditions we have $a b (1+a+b) \geq 3$. If $a+b < 2$, then $a b \leq \left(\frac{a+b}{2}\right)^2 < 1$ and thereby $a b (1+a+b) < 3$. So the implication $a b (1+a+b) \geq 3 \Rightarrow a+b \geq 2$ is indeed true.