The Schroeder-Bernstein Theorem holds in categories where every endomorphism is an isomorphism. More precisely, if every endomorphism is an isomorphism and there are morphisms $f:A\rightarrow B,g:B\rightarrow A$, then $A$ and $B$ are isomorphic.

It turns out that the category of ultrafilters has no non-trivial endomorphisms and hence the Schroeder-Bernstein Theorem holds for the category of ultrafilters.

To be precise, the objects in the category of ultrafilters are pairs of the form $(X,\mathcal{U})$ where $X$ is a set and $\mathcal{U}$ is an ultrafilter on the set $X$. If $(X,\mathcal{U}),(Y,\mathcal{V})$ are objects in the category of ultrafilters, then a premorphism from $(X,\mathcal{U})$ to $(Y,\mathcal{V})$ is a function $f:X\rightarrow Y$ such that if $R\subseteq Y$, then $R\in\mathcal{V}$ iff $f^{-1}[R]\in\mathcal{U}$. Now let $\mathcal{A}_{(X,\mathcal{U}),(Y,\mathcal{V})}$ be the set of all premorphisms from $(X,\mathcal{U})$ to $(Y,\mathcal{V})$. Then define an equivalence relation $\simeq$ on $\mathcal{A}_{(X,\mathcal{U}),(Y,\mathcal{V})}$ so that $f\simeq g$ iff $\{x\in X|f(x)=g(x)\}\in\mathcal{U}$. Then the morphism between $(X,\mathcal{U})$ and $(Y,\mathcal{V})$ are the equivalence classes in $\simeq$. The composition of morphisms is ordinary composition of functions. Every endomorphism in the category of ultrafilters is the identity mapping.

In fact, the fact that the category of ultrafilters satisfies the Schroeder-Bernstein theorem is one reason why the Rudin-Keisler ordering on ultrafilters is well known while the actual category of ultrafilters is less well known:

Take note that $\mathcal{U}\leq_{RK}\mathcal{V}$ if and only if there is a morphism from $\mathcal{V}$ to $\mathcal{U}$. Since the category of ultrafilters satisfies the Schroeder-Bernstein Theorem, if $\mathcal{U}\leq_{RK}\mathcal{V}\leq_{RK}\mathcal{U}$, then $\mathcal{U}$ and $\mathcal{V}$ are isomorphic ultrafilters.
Therefore the Rudin-Keisler ordering on ultrafilters is a satisfactory ordering on the class of ultrafilters because the category of ultrafilters satisfies the Schroeder-Bernstein Theorem (and because the ultrafilters have no non-trivial endomorphisms also).

The Rudin-Keisler ordering has been generalized to other kinds of objects such as filters and Boolean ultrapowers. However, the generalized Rudin-Keisler orderings on the category of filters and the category of Boolean ultrapowers respectively does not satisfy the Schroeder-Bernstein Theorem (presumably), so the generalization of the Rudin-Keisler ordering are less satisfactory in these categories (and yes filters and Boolean ultrapowers do form categories).