In #2,3,4 you have one element which squares to the identity, and two elements which square to that element. That is, your group is of the form
{e,x,x2,x3}, where
x2 is the element that squares to the identity and
x,x3 both square to
x2. Since you're forced to have
a=e, it is left open to relabel
x,x2,x3 as
a,b,c. But since we could let
y=x3 and the group would be
{e,y3,y2,y}={e,y,y2,y3} it makes no difference whether an element is x or x³, so the only element where the choice "matters" is
x2, hence three different-looking tables (if you put a,b,c,d in that order).
In #1 every element of your group squares to the identity, and so it's of the form
{e,x,y,xy}. But notice that you could let, say,
p=x,q=xy, and then you'd have
{e,p,pq,q}={e,p,q,pq}, so nothing changes no matter what you call any of the elements, hence only one different-looking table (if you put a,b,c,d in that order).