Here is an easy argument as to why the answer is $\omega_1$.
$\sf ZF$ (and in fact $\sf Z$) proves that given any two well-ordered sets, one is isomorphic to a unique initial segment of the other. That is, any two well-ordered sets are comparable.
By definition, $\omega_1$ is the smallest order type of an uncountable well-order. So if we prove that $\omega_1\setminus\omega$ is uncountable, then $\omega_1$ is isomorphic to an initial segment of it, but since it is also a subset of $\omega_1$, this initial segment cannot be a proper initial segment, as that would imply $\omega_1$ is isomorphic to a proper initial segment of itself which contradicts the uniqueness part of the above theorem.
But now it's easy. If $\omega_1\setminus\omega$ is countable, then $\omega_1$ is the union of two countable sets, which is indeed countable. That's not true, so the order type is $\omega_1$.
Finally, if you want an explicit argument, which you probably do, note that just like you can add an element to the bottom of $\omega$ and not change the order type, you can also add an $\omega$ sequence to the bottom of the ordinal $\omega\cdot\omega$ without changing the order type. Apply that sort of bijection on the initial segment that is $\omega\cdot\omega$ and the identity elsewhere, and you get your order isomorphism.