Does convergence for Cauchy sequence fail only when the limit is not in the domain?

This can be viewed as "the only kind of failure mode", since any non-complete metric space can be viewed as a subspace of a complete metric space. For more details, see e.g. this question and its answers.

Yes, it is the only kind, in the following sense: If $(X,d)$ is a metric space and if $(a_n)_{n\in\mathbb N}$ is a Cauchy sequence of elements of $X$ which does not converge, then there is a metric space $(Y,d^\star)$, such that $Y\supset X$ and that $(\forall x,x'\in X):d^\star(x,x')=d(x,x')$ and the sequence $(a_n)_{n\in\mathbb N}$ converges in $Y$.

A sequence $(x_n)_n$ in a metric space is not convergent iff the set of limit points of the sequence is not a singleton. So iff it is empty or has more than one element.

If $(x_n)_n$ is a Cauchy-sequence then it can be proved that there is no more than one element in that set, so if the sequence is not convergent then the set of limit points must be empty.

This can always be "repaired" by completion.