Essentially I was wondering if the quotient of an algebraic group $G$ by its identity component $G^0$ is necessarily always parabolic.
My argument:
This seems right since $G^0$ is a closed subgroup of $G$, therefore the quotient $G/G^0$ makes sense. Moreover its points are the fibers above $G^0$ hence is a finite quasi-projective variety. Thus it must be isomorphic (as a variety) to $\sqcup_{\text{finite}} \mathbb{A}^0$.
Note: The latter is complete? Since, for any variety $X$, $\sqcup_{\text{finite}} \mathbb{A}^0 \times X \cong X$ (as varieties).
You are correct that the identity component $G^0$ is a parabolic by your definition. Your argument is correct up until you say that $G/G^0 \times X \simeq X$.
Instead, to see that $G/G^0$ is complete observe that it is a finite group and so discrete as a topological space. This means that the irreducible components of $G/G^0$ are points. It's trivial that a point is complete and it's a good exercise to try and prove that a space is complete if and only if its irreducible components are complete.