That conclusion that $Q/Z(Q) \cong V$ is founded on that $|Q/Z(Q)| = 4$. Thus $Q/Z(Q)$ is isomorphic to either $\mathbb Z_4$ or $V$. (Why? These group operations are the only valid ones on a set of four elements. To see this, try writing out a group table for a group of four elements just from the axioms. You will end up with exactly two possibilities up to isomorphism)
So our task is to figure out which group $Q/Z(Q)$ is isomorphic to. We know that order is preserved through isomorphism, and $\mathbb Z_4$ has two elements of order 4 while $V$ has none. Thus observing that $Q/Z(Q)$ has all elements of order $\le 2$ shows that it cannot be isomorphic to $\mathbb Z_4$ and we are done.
The abelianess without any appeal to orders of elements, Cauchy's Theorem or Lagrange's or type of group: assume that $G$ has $4$ elements and is not abelian. Then we can find two non-identity elements $a,b$ that do not commute, so $ab \neq ba$. Note that this implies $ab \notin \{e,a,b\}$. Because $G$ is closed under the group operation we must have $G=\{e,a,b,ab\}$. Now $ba$ belongs to this set. Since $a$ and $b$ do not commute $ba \notin \{e,a,b\}$. Hence we must have $ab=ba$, a contradiction. So $G$ is abelian after all.
OK, so we have $\{e,a,b,ab\}$ is an abelian group of $4$ elements, so each of them does not equal one of the others. So were is $a^2$? Assume for the moment that $a^2 \neq e$. Can $a^2=a$? No, since then it would follow $a=e$. Can it be $a^2=b$? Yes, and it that case $ab=a^3$ and the group looks like $\{e,a,a^2,a^3\}$. Observe that $a^4$ must be the identity since if $a^4=a$, $a^4=a^2$ or $a=a^3$, the set reduces to less than $4$ elements. So the group is cyclic of order $4$ generated by $a$, in this case.
Similarly, if $b^2=e$, we would arrive at the group $\{e,b,b^2,b^3\}$, which is of course again cyclic of order $4$ and isomorphic to the one we already found.
We are left with the case where $a^2=e=b^2$. But then (using abelianess) $(ab)^2=abab=aabb=a^2b^2=e.e=e$ and now all elements of the group have order $2$. This one is isomorphic to $C_2 \times C_2$, a direct product of two groups of order $2$, also called the Klein $4$-group $V_4$.
So you see in this small case everything can still be figured out without using more sophisticated theorems.
Best Answer
Basically any property that is preserved by group isomorphisms will do. This includes:
Another method (for finite groups) is to look at character tables. If two groups have different character tables, they can't be isomorphic. Notice that the converse is false: two groups can have the same character table without being isomorphic (If I remember correctly $D_8$ and $Q_8$ are a counterexample when considering the character table for the complex irreducibel representations, but correct me if I'm wrong!)
In practice, one can look at special subgroups like the center and normalisator of a group, as these are often easier to understand than the entire group.