The class of $\aleph$ numbers is the same class of cardinals that you know in a model of ZFC. Namely, initial ordinals. The definitions are exactly the same. Furthermore by definition the $\aleph$ cardinals are ordinals, so the correspond to well ordered sets.
On the other hand, if $A$ is not a well-orderable set, then $|A|$ corresponds to the set $$\{B\mid \exists f\colon A\to B\text{ a bijection}\land\operatorname{rank}(B)\text{ is minimal}\}$$
Where the $\operatorname{rank}$ operator is the von Neumann rank of $B$. This set is not an ordinal, clearly, and it may lack any internal structure.
The class of cardinals, therefore, is combined from two parts:
The $\aleph$ numbers which are "ordinal which cannot be put in bijection with any of its elements".
We can see that the $\aleph$ numbers do not form a set directly, suppose that they would, then there was an ordinal $\gamma$ such that the set of $\aleph$ has von Neumann rank $\gamma$. In particular all of its elements have rank $<\gamma$. Let $\kappa$ be the first ordinal above $\gamma$ such that $\kappa$ is not in bijection with any of its elements, then $\kappa$ is an $\aleph$, but its von Neumann rank is $\kappa>\gamma$ in contradiction.
Cardinals of sets which are not well-orderable. These are described as sets $A$ such that "Every two members of $A$ have a bijection between them, all the elements of $A$ have the same von Neumann rank, and no set of lower rank has a bijection with any element of $A$, and if there is a $B$ of the same von Neumann rank as a member of $A$, and they are in bijection then $B$ is an element of $A$ as well"
Yes, it is a bit clumsy and unclear, but set theory without choice may get like that often.
It is immediate that the class of cardinals is a proper class since it contains all the $\aleph$-cardinals. Much like in ZFC the cardinals make a proper class, the arguments carry over in this case as well.
Lastly, you cannot prove that a power set of a well-ordered set is well-ordered because if the axiom of choice fails this is simply not true. Furthermore, $A$ itself is a class, as it contains elements of unbounded rank, so we need to be more careful with "the union over $A$" as it is not a set as well, that is $V$ itself is a class.
As $V$ is a class its power "set" is not a set and does not exist, and as I remarked power sets of a well-ordered set need not be well-orderable.
See also:
- Defining cardinality in the absence of choice
- There's non-Aleph transfinite cardinals without the axiom of choice?
- How do we know an $ \aleph_1 $ exists at all? (this asserts that $\aleph_1$ exist, even without choice, and the argument carries over to high cardinals)
There's no clever arithmetic here.
Take a set $a$, and let $u=a\times\omega$. Note that $u+u=a\times\omega\times 2$, so $2u=u$. Now let $v=\aleph(u)$, the Hartogs number of $u$, then $2u=u<u+v$, therefore $u<v$, but that means that $u$ injects into an ordinal, and since $a$ injects into $u$ we get that $a$ injects into an ordinal and can be well-ordered.
Best Answer
If I understand the problem correctly, it depends on your definition of cardinal. If you define the cardinals as initial ordinals, then your argument works fine, but without choice you cannot show that every set is equinumerous to some cardinal. (Since AC is equivalent to every set being well-orderable.)
On the other hand, if you have some definition which implies that each set is equinumerous to some cardinal number, then without choice you cannot show that any two sets (any two cardinals) are comparable. (AC is equivalent to: For two sets $A$, $B$ there exists either an injective map $A\to B$ or an injective map $B\to A$. It is listed as one of equivalent forms if AC at wiki.)