Q:

Does having exactly one of each element in every row and column of a Cayley table ensure that it’s a group

A group must have a Cayley table in which each row and column has one and only one of each element. This can be proved by considering the opposite: suppose one row of a set’s Cayley table did not contain a particular element. That is, let AX not equal B, whatever X is. Let C be A’s inverse, such that CA = I. We easily reach a contradiction of associativity with ACB. (A*C)B = IB = B. On the other hand, A*(CB) = AX, does not equal B. Hence, each row and column of a group's Cayley table must contain exactly one of each element.
Two questions then arise:
Does a Cayley table in which each row and column has exactly one of each element guarantee that it is a group?
For a given number of elements, how many possible non-isomorphic groups are there? For instance, there is only one group with three elements. I think there are 4 groups with four elements.

A:

No!  Such a Cayley table will not necessarily satisfy the associative law, which is an absolutely critical property in group theory.  
You can easily construct a counterexample on 5 elements.

a b c d e
b c d e a 
c a e b d
d e b a c
e d a c b

Every symbol occurs exactly once in every row or column, and $a$ is the identity element, but this is not associative: $(b*b)*b = a$ while $b*(b*b) = d$.  (The multiplication $xy$ is defined by looking up row $x$ and column $y$.)
In general, Cayley tables are not a good way of constructing groups. Even though Cayley tables are usually presented in beginner texts due to their resemblance to the familiar addition/multiplication tables, the associative property is difficult to check in a Cayley table. Instead, groups are usually constructed in terms of some operation which is already known to be associative, such as function composition, integer addition or multiplication, or (once you get off the ground) previously constructed groups.

