There is an ambiguity in this kind of question, where you refer to ZFC-A for an axiom A, since what the theory is that you refer to depends on the particular way that you axiomatize ZFC. For example, different axiomatizations that are equivalent in the presence of axiom A might no longer be equivalent when you omit A, and so it could be unclear what is meant by ZFC-A.

This kind of issue was the main point of my paper with Gitman and Johnstone, to which you refer: What is the theory ZFC without power set? In this paper, we prove that a variety of surprising statements are consistent with ZFC-power, if one axiomatizes this theory via the replacement axiom, but a better theory is obtained if one axiomatizes the theory via the collection axiom, plus separation (the point being that this is not equivalent to replacement when one lacks the power set axiom). Our summary conclusion is that one should really use collection+separation rather than replacement, when omitting power set, since otherwise so much goes wrong. Thus, when properly axiomatized, I wouldn't agree that $\omega_1$ can be singular in ZFC - powerset. That situation is only consistent with an improper formulation of the theory.

Regarding your specific question at the end, one doesn't need the replacement axiom to prove that $\omega_1$ is regular; this can be proved in the Zermelo theory plus choice, assuming that $\omega_1$ exists. First, one forms the set of relations on $\omega$ that are well-orders. These can be grouped into equivalence classes under the order-isomorphism relation. Now, by AC, one can pick a representative from each class. If $\omega_1$ is singular, then one gets a countable sequence of such representatives of order type unbounded in $\omega_1$. And one can now glue these orders together to make a single countable order type at least as large as $\omega_1$, a contradiction.