What do you mean by an irreducible Cartier divisor? Assume that $S$ and $D$ are regular, that $\mathcal X$ is normal and has finite inertia, and that $f^*D = kE$ for a reduced divisor. Also assume that $\mathcal X$ is tame in codimension 1. Then the induced morphism $\mathcal X\to \sqrt[k]{D/S}$ is proper, because both stacks are proper over $S$. It is also birational. I claim that is representable in codimension 1; this follows from the fact that $\mathcal X$ is ramified of degree $k$ at the generic point of each irreducible component of $D$ (this can be done, for example, by taking the strict henselization of $S$ at the generic point of such a component, thus reducing to the case that $S$ is an henselian trait, which is easy, using the tameness hypothesis). Thus $\mathcal X\to \sqrt[k]{D/S}$ is a proper morphism with finite fibers, $\mathcal X$ is normal, $\sqrt[k]{D/S}$ is regular, and is an isomorphism in codimension 1. By purity of branch locus, it must be étale; and then it must be an isomorphism.

I think that all of the hypotheses are necessary. For example, already when $D$ is a nodal curve on a smooth surface $S$ there are counterexamples: there is a smooth stacks having $S$ as its moduli space, which is ramified of order $k$ along $D$ (this is different from $\sqrt[k]{D/S}$, because the latter is singular). For example, when $D$ is the union of two smooth curves intersecting transversally, you take the fiber product of the root stacks of the two curves. There are also counterexamples when $\mathcal X$ is not normal, or when it is not tame.

What do you mean by an irreducible Cartier divisor? Assume that $S$ and $D$ are regular, that $\mathcal X$ is normal and has finite inertia, and that $f^*D = kE$ for a reduced divisor $E$ on $\mathcal X$. Also assume that $\mathcal X$ is tame in codimension 1. Then the induced morphism $\mathcal X\to \sqrt[k]{D/S}$ is proper, because both stacks are proper over $S$. It is also birational. I claim that is representable in codimension 1; this follows from the fact that $\mathcal X$ is ramified of degree $k$ at the generic point of each irreducible component of $D$ (this can be done, for example, by taking the strict henselization of $S$ at the generic point of such a component, thus reducing to the case that $S$ is an henselian trait, which is easy, using the tameness hypothesis). Thus $\mathcal X\to \sqrt[k]{D/S}$ is a proper morphism with finite fibers, $\mathcal X$ is normal, $\sqrt[k]{D/S}$ is regular, and is an isomorphism in codimension 1. By purity of branch locus, it must be étale; and then it must be an isomorphism.

I think that all of the hypotheses are necessary. For example, already when $D$ is a nodal curve on a smooth surface $S$ there are counterexamples: there is a smooth stacks having $S$ as its moduli space, which is ramified of order $k$ along $D$ (this is different from $\sqrt[k]{D/S}$, because the latter is singular). For example, when $D$ is the union of two smooth curves intersecting transversally, you take the fiber product of the root stacks of the two curves. There are also counterexamples when $\mathcal X$ is not normal, or when it is not tame.

What do you mean by an irreducible Cartier divisor? Assume that $S$ is regular (you can weaken this) that $D$ is reduced, that $\mathcal X$ has finite inertia, and that $f^*D = kE$ for a reduced divisor. Then the result is true. The induced morphism $\mathcal X\to \sqrt[k]{D/S}$ is proper, because both stacks are proper over $S$; and from the hypothesis on $D$ it follows that $\mathcal X\to \sqrt[k]{D/S}$ is representable (this is an exercise, I can add more details if you'd like; the point is that $\mathcal X$ must have ramification index $k$ at each point of $E$). Then the result follows immediately.

What do you mean by an irreducible Cartier divisor? Assume that $S$ and $D$ are regular, that $\mathcal X$ is normal and has finite inertia, and that $f^*D = kE$ for a reduced divisor. Also assume that $\mathcal X$ is tame in codimension 1. Then the induced morphism $\mathcal X\to \sqrt[k]{D/S}$ is proper, because both stacks are proper over $S$. It is also birational. I claim that is representable in codimension 1; this follows from the fact that $\mathcal X$ is ramified of degree $k$ at the generic point of each irreducible component of $D$ (this can be done, for example, by taking the strict henselization of $S$ at the generic point of such a component, thus reducing to the case that $S$ is an henselian trait, which is easy, using the tameness hypothesis). Thus $\mathcal X\to \sqrt[k]{D/S}$ is a proper morphism with finite fibers, $\mathcal X$ is normal, $\sqrt[k]{D/S}$ is regular, and is an isomorphism in codimension 1. By purity of branch locus, it must be étale; and then it must be an isomorphism.

I think that all of the hypotheses are necessary. For example, already when $D$ is a nodal curve on a smooth surface $S$ there are counterexamples: there is a smooth stacks having $S$ as its moduli space, which is ramified of order $k$ along $D$ (this is different from $\sqrt[k]{D/S}$, because the latter is singular). For example, when $D$ is the union of two smooth curves intersecting transversally, you take the fiber product of the root stacks of the two curves. There are also counterexamples when $\mathcal X$ is not normal, or when it is not tame.