It is easy to solve $x^3-1=0$ by having the form $(x-1)(x^2+x+1)=0$
The $3$ roots are: $x=1$, $x=\frac{-1\pm i\sqrt 3}{2}$.
But, solving the equation $x^3+1=0$ by using the similar approach of using the standard approach to solving the quadratic equation & the complex roots does not yield.
$x^3+1=0\implies x^3 -1=-2 \implies (x-1)(x^2+x+1)=-2$ leads to $x-1=-2\implies x=-1$.
The solution $x^2+x+1=-2$ will lead to $x^2+x+3=0$ with roots $x =\frac{-1\pm i \sqrt{11}}{2}$.
But, the answer is $x =-1, \frac{1\pm i \sqrt{3}}{2}$.
Your error is in the line where you say that
This is not true. In general, $ab = -2$ does not imply that $a$ or $b$ are in the set $\{1, 2, -1, -2\}$; for example, we could have $a = \pi$ and $b = -2/\pi$. You seem to have mixed this up with a property of zero. The valid implication is that
$$ab = 0 \implies a = 0 \text{ or } b = 0$$
for complex numbers $a, b$. Using this observation with the factorization
$$x^3 + 1 = (x + 1)(x^2 - x + 1)$$
will lead to a viable solution.