I believe I've found a third and fourth group of degree four. Tell me how wrong I am.

Let's call your first group $G = \{e,a,b,c\}$, and your second group $H = \{1,x,y,z\}$.

Note that $b^2 = a$, so we may re-write $G = \{e,b^2,b,c\}$. Next, note $b^3 = b^2b = ab = c$, so we have $G = \{e,b^2,b,b^3\} = \{e,b,b^2,b^3\}$ making it clear that $b$ generates $G$.

This suggests the map $\Bbb Z_4 \to G$:

$0 \mapsto e\\1 \mapsto b\\2 \mapsto b^2 = a\\3 \mapsto b^3 = c$

and to verify this is an isomorphism, you need only verify:

$k + m(\text{ mod }4) \mapsto b^{k+m(\text{ mod }4)} = b^kb^m$ for $k, m \in \{0,1,2,3\}$

A similar analysis holds with $H$ using $x$ as a generator.

You're correct in some sense in that your groups are indeed distinct products on a set of four elements, but "re-naming" the elements leads to the same kind of group structure (cyclic) which is what the notion of isomorphism is meant to capture (the same in all important algebraic ways, but not necessarily identical sets).