A bit of a response to your "Commentary":

As you point out, the failure of your construction to hit all $\mu_n$-gerbes is governed by the exact sequence
$H^1(X, \mathbb{G}_m) \to H^2(X, \mu_n) \to H^2(X,\mathbb{G}_m)[n] \to 0$ so answering your question is related to producing torsion $\mathbb{G}_m$-torsors.

The question of doing so has been studied as part of the theory of Brauer groups:

Let $Br(X)$ ("Brauer group") denote the group of *Azumaya algebras*, which are a generalization of the central simple algebras over a field (that is the classical Brauer group).

Let $Br'(X)$ ("Cohomological Brauer group") denote the *torsion part* of $H^2(X, \mathbb{G}_m)$.

In the case of $X = Spec k$, $k$ a field, the equivalence of these two groups is classical and they can be computed in various cases of number-theoretic interest (e.g., number fields/local fields/finite fields). In this case, $H^1(X, \mathbb{G}_m)=0$ by Hilbert's Theorem 90, and yet there are plenty of examples where $Br(X)$ is very much non-trivial. Grothendieck studied the relation between $Br(X)$ and $Br'(X)$ in general in Dix Exposes. The upshot is that there is an injective map $Br(X) \to Br'(X)$ and it is an isomorphism in reasonable cases (e.g., I think $X$ quasi-projective over a field). (See Dix Exp, or Ch. IV of Milne's "Etale Cohomology".)

I won't say more about the general picture, but I'll work out in detail the simple case of $X = Spec \mathbb{R}$. In this case, $Br(Spec \mathbb{R}) = \mathbb{Z}/2\mathbb{Z}$ generated by the class of the usual quaternions, viewed as a central simple algebra over $\mathbb{R}$. We can give a geometric description of the resulting $\mathbb{G}_m$-torsor:

Start with the smooth plane conic $C = Proj \mathbb{R}[x,y,z]/(x^2+y^2+z^2)$. It's a smooth genus $0$ curve, but has no $\mathbb{R}$-points and so is not isomorphic to $\mathbb{P}^1_{\mathbb{R}}$. However, after base-change to $\mathbb{C}$ it attains a point and so becomes isomorphic to $\mathbb{P}^1_{\mathbb{C}}$; such a Galois-twisted form of $\mathbb{P}^n_{\mathbb{C}}$ is known as a Brauer-Severi variety and the elements of the Brauer-group (of a field) can also be thought of as corresponding to them (the group structure is then a bit strange). Since $Aut(\mathbb{P}^n) = PGL_{n+1}$, these correspond to $PGL_{n+1}$-torsors and the relation to $\mathbb{G}_m$-torsors is via the exact sequence for $PGL_{n+1}=GL_{n+1}/\mathbb{G}_m$. So, a $T$-point of the corresponding $\mathbb{G}_m$-torsor for a $\mathbb{R}$-scheme $T$ consists of the following data:

It is a rank $2$ vector bundle $V$ over $T$, together with an isomorphism of $T$-schemes $C_T \simeq \mathbb{P}(V)$ where $C_T = C \times_{Spec \mathbb{R}} T$ is the pullback of our genus $0$ curve to $T$, and $\mathbb{P}(V)$ is the associated projective space (here $\mathbb{P}^1$) bundle of our vector bundle $V$.

Why is this a $\mathbb{G}_m$-gerbe? Well, $\mathbb{P}(V) \simeq \mathbb{P}(V')$ iff $V$ and $V'$ differ by tensoring by a line bundle. The gerbe is non-trivial since it has no $\mathbb{R}$-points, since $C$ itself is not isomorphic to projective space. It has $\mathbb{C}$-points because the base-change is isomorphic to projective space.