**Update**: I removed what I thought was unecessary and tried to be more straightforward in the hope to get an answer.

**Context:**
Suppose I have a $\mathbb{G}_m$-gerbe $\mathcal{G}$ over a scheme $X$ with the fppf (or lisse-etale) topology. Because $\mathcal{G}$ is a $\mathbb{G}_m$-gerbe, there is an fppf (or etale) morphism of schemes $U \xrightarrow{i} X$ trivializing $\mathcal{G}$. So we have canonical isomorphism $\big(B\mathbb{G}_m\big)_U \cong U \times_X \mathcal{G}$ of $\mathbb{G}_m$-gerbes over $U$ and a morphism of stacks $\pi: \big(B\mathbb{G}_m\big)_U \to \mathcal{G}$.

**What I want to know:**
If $\mathcal{F}$ is a quasi-coherent sheaf over $\mathcal{G}$ we often read the simplifying argument that because the question is etale local on $X$ one can assume $\mathcal{G} = B\mathbb{G}_m$. What is the justification for this?

**Motivation for the question:**
I'm interested to prove some equivalence of categories of quasi-coherent sheaves with some extra property defined over some $\mathbb{G}_m$-gerbes and I know how to prove the result for $B\mathbb{G}_m$ (i.e. I know how to prove the equivalence once both gerbes have been trivialized over a common $X$-scheme). I would like to explain why this equivalence also holds over the non-trivial gerbes. I can be more explicit about this part if needed.

**Attempt to answer the question**
I have a $\mathbb{G}_m$-gerbe $\mathcal{G}$ over $X$. There is an etale surjective morphism $U \to X$ such that $U \times_X \mathcal{G} \cong (B\mathbb{G}_m)_U$.

Now I have an atlas $s: U \to (B\mathbb{G}_m)_U$ inducing an atlas $\pi \circ s : U \to \mathcal{G}$, where $\pi$ is the projection onto $\mathcal{G}$.

I then have a presentation of my gerbe

$$ U \times_\mathcal{G} U \times_\mathcal{G} U \to U \times_\mathcal{G} U \to U \xrightarrow{\pi \circ s} \mathcal{G} $$

and a sequence

$$ (B\mathbb{G}_m)_U \times_\mathcal{G} (B\mathbb{G}_m)_U \times_\mathcal{G} (B\mathbb{G}_m)_U \to (B\mathbb{G}_m)_U \times_\mathcal{G} (B\mathbb{G}_m)_U \to (B\mathbb{G}_m)_U \xrightarrow{\pi} \mathcal{G}. $$

If $\mathcal{F}$ is a quasi-coherent sheaf then on the atlas $U$ I get a quasi-coherent sheaf $(\pi \circ s)^* \mathcal{F}$ together with an isomorphism $\sigma: pr_1^*(\pi \circ s)^* \mathcal{F} \to pr_2^*(\pi \circ s)^* \mathcal{F}$ over $U \times_\mathcal{G} U$ which satisfies the cocycle condition over $U \times_\mathcal{G} U \times_\mathcal{G} U$.

Now I believe that we have

$$ s_*(\pi \circ s)^* \mathcal{F} = \pi^* \mathcal{F} $$

over $(B\mathbb{G}_m)_U$.

So I have this quasi-coherent sheaf together with an isomorphism $(s \times s)_*\sigma$ over $(B\mathbb{G}_m)_U \times_\mathcal{G} (B\mathbb{G}_m)_U$ (where $s \times s$ is the universal map induced by $s$) which satisfies the cocycle condition over triple fiber product.

This would be what I would call the descent data for the sheaf $\mathcal{F}$ and one could define the category of descent $QCoh \big((B\mathbb{G}_m)_U \to \mathcal{G} \big)$ with the objects similar to the one described above (obviously not necessarely assuming that it comes from a sheaf on the gerbe $\mathcal{G}$).

**Question:**
Now this descent argument doesn't technically reduce to the case of $B\mathbb{G}_m$ but to a descent data of a sheaf on it. Is this what is usually intended?

**Relation to the motivation:**

Assume that what I said was correct, then would it be it exact to say the following: that to prove a given equivalence of categories as in the motivation I could equivalently have to prove that the categories of descent over the same $B\mathbb{G}_m$ are equivalent? Then in order to do that it would suffice to show that the given categories of quasi-coherent sheaves + extra property over $B\mathbb{G}_m$ are equivalent? (To justify the last part I think that if I can prove that the $QCoh(B\mathbb{G}_m)$ + property are equivalent, then the descent data would almost automatically correspond to one another through this equivalence? My property is some $\mathbb{G}_m$-action on the sheaves so it won't cause any problem on the descent data)