Firstly, does the family of all non-isomorphic non-Abelian groups have a well defined cardinality? How about the family of all non-isomorphic Abelian groups? If they are both defined, how do they compare?
Does the statement "there are more non-Abelian groups than Abelian groups" make any mathematical sense at all?
I am aware of a related question but my question should be different.
On
If we restrict ourselves to finitely presented abelian or nonabelian groups, there are countably many nonisomorphic groups in each set.
Since there are countably many finite group presentations, we need only find a countable subset of each. Indeed, the free groups, $F_n$ ($n\ge 2)$, and the free abelian groups, $\mathbb{Z}^n$, will do. These are pairwise non-isomorphic (which seems obvious, but all the proofs I know are mildly technical).
We can even find the same result among finite groups: the cyclic groups, $\mathbb{Z}/n\mathbb{Z}$, and the symmetric groups $S_n$ ($n \ge 3$) are pairwise non-isomorphic (they have different orders) and there are countably many.
In the case of finite simple groups, there is a sense in which your statement holds: according to the classification of finite simple groups there is only one infinite family of finite simple abelian groups (cyclic groups of prime order), but two (or more, depending on how you count) families of nonabelian groups: the alternating groups $A_n$ ($n \ge 5$) and the simple groups of Lie type.
No, being a proper class, it does not have a cardinality.
No: same reason.
N/A
Maybe, but not in the sense of cardinality. You would have to be more specific about the context you are working in.