If a set $A$ has the same cardinality as an ordinal $\alpha$, then there exists a bijection $f:\alpha\to A$, so $A$ is indexed by $\alpha$ and hence well-ordered. Therefore a choice function $g:\mathcal{P}(A)\to A$ exists.
Therefore, if Axiom of Choice(AC) is false, then there must exist a set $A$, such that no ordinals have the same cardinality of $A$.
So the equivalence class of sets with the same cardinal is larger if AC is false than if AC is true.
As a result, the truth or falsity of the Continuum hypothesis(CH) needs to be considered separately with AC and without AC.
Is the reasoning above correct? and is there really two cases to consider for CH?
EDIT: Check my idea about "larger": Let two sets $X\sim Y$ iff they have the same cardinality. $\sim$ is an equivalence relation. Denote the class of all the equivalence classes under AC $S_1$, and the class of all the equivalence classes without AC $S_2$. Clearly $S_1\subseteq S_2$, since $S_2$ has elements not in the set Ord(ordinals). This is really not formal, since it depends on the model we use.
Best Answer
It is a bit misleading to say that it is "larger" (how do you compare them ?) but you are right about the idea, and the formal statement : AC is indeed equivalent to "for all $A$, there is an ordinal $\alpha$ and a bijection $\alpha\to A$".
You are also right about CH, with AC all the statements that could reasonably called CH are equivalent, while this is no longer true without AC.
To give an example of a statement that amused me, under the Axiom of Determinacy (AD - it's incompatible with AC, but under some large cardinal hypothesis, if ZF is consistent, then so is ZF+AD), any subset of $\mathbb{R}\sim 2^\omega$ is either countable, or has the same size as $\mathbb{R}$. One could say that this means that CH is satisfied under AD.
On the other hand though, $\mathbb{R}$ is not equipotent to $\omega_1$, in fact the two have incomparable cardinalities. One could say that this means that CH fails under AD.
Without AC, you have to be more careful about what you mean with CH, and one may even say that CH is not really meaningful without AC.