Cohomology of quotient stack Let $X$ be an algebraic variety over $\mathbb{C}$ (the ground field is not important but this makes things easier I think) and $G$ an algebraic group acting over it. Let's say we know that there's a normal subgroup $K \cong \mathbb{G}_m$ such that $K$ acts trivially on $X$ and $G/K$ acts in a free way on $X$.
I think one could put less restrictive hypothesis on $K$ actually: maybe reductive should be enough.
I'm denoting as $\mathcal{X}=[X/G]$ the quotient stack. Is it always true that $$H^*(\mathcal{X}) \cong H^*(X/(G/K)) \otimes H^*(BK) $$ and $$H^*_c(\mathcal{X}) \cong H^*_c(X/(G/K)) \otimes H^*_c(BK)  $$
(everything is meant to be with $\overline{\mathbb{Q}_{\ell}}$ coefficients.
What should be a possible counterexample?
 A: Let $G$ be a linear algebraic group, and $K\subset G$ a closed normal subgroup with quotient $L = G/K$. So all three groups are linear algebraic. Further, assume $L$ is connected.
Let $X$ be a variety with $G$-action such that $K$ acts trivially, $L$ acts freely and we have a quotient $X/L$.
Step 1:  The canonical map $BG \to BL$ is a fiber bundle with fiber $BK$. As $L$ is connected the mondromy action on the cohomology of the fibers, namely $H^*(BK)$, is trivial. Further, the Leray-Serre spectral sequence associated to this fiber bundle degenerates at $E_2$ because both $BK$ and $BL$ have cohomology concentrated in even degrees.
Note: The claim about odd vanishing in cohomology holds because $K$ and $L$ are linear algebraic groups and we are using torsion free coefficients (it is ultimately due to the Bruhat decomposition). To emphasize: no connected assumptions are needed for this vanishing, just torsion free coefficients and linear algebraic groups.
Step 2:  We have a pullback square (this claim should be checked - I have not worked out the details carefully):
$\require{AMScd}$
\begin{CD}
[X/G] @>{}>> BG\\
@VVV @VVV\\
X/L @>{}>> BL
\end{CD}
The monodromy action on the cohomology of the fibers $H^*(BK)$ for the bundle $[X/G] \to X/L$ is the pullback of the monodromy action in $BG \to BL$. Hence, it is trivial.
Further, the elements of $H^*(BK)$ are permanent cycles in the Leray-Serre spectral sequence of $[X/G] \to X/L$ because they are so for that of $BG \to BL$. This gives the required claim:
$H^*([X/G]) \cong H^*(BK) \otimes H^*(X/L)$.
Note: even assuming I haven't done something completely dumb above, this says nothing about ring structure. Further, the requirement that $L$ act freely on $X$ should be unnecessary - replace $X/L$ by $[X/L]$ (again, modulo me not being delusional).
The point (assuming the pullback square is correct) at the heart of the matter is that $BG \to BL$ is the universal example of a fiber bundle with structure group $G$ in which the $G$-action on the fibers factors through $L$ (i.e., $K$ acts trivially).
