I'm curious if there are any cayley tables on a finite amount of elements that satisfy the axioms of a) closure, b) identity, and c) inverse, but that for at least one triple of elements do not satisfy the associative property, and so thus the set is not a group.
I'm also wondering if there are any in which every element has a unique inverse but a group is still not formed because associativity is not held.
On
Consider the following Cayley Table of order $5$:
$$\begin{array}{c|ccccc} \ast & 1 & 2 & 3 & 4 & 5\\ \hline 1 & 1 & 2 & 3 & 4 & 5\\ 2 & 2 & 1 & 3 & 3 & 5\\ 3 & 3 & 3 & 1 & 4 & 5\\ 4 & 4 & 3 & 4 & 1 & 5\\ 5 & 5 & 5 & 5 & 5 & 1 \end{array}$$
We have that $\{1,2,3,4,5\}$ is closed under $\ast$, $1$ is the identity element and every element is the unique left/right inverse of itself (see the diagonal of the table).
Moreover, this operation $\ast$ is commutative too, but not associative:
$$(2\ast 3)\ast 4 = 3\ast 4 = 4 \ne 3 = 2\ast 4 = (2\ast(3\ast 4)).$$
What's interesting is that one such structure, that
satisfies all group axioms except associativity,
is not necessarily even a quasigroup. Indeed in my example, as in the first one of two in the answer by @David, not all the rows (columns) of the Cayley Table are bijections of the underlying set: such structures are indeed not cancellative and have no division (see definitions 5 to 9 in this paper).
For the table $$\matrix{e&a&b\cr a&e&e\cr b&b&e\cr}$$ we have $$(ab)a=ea=a\quad\hbox{but}\quad a(ba)=ab=e\ .$$
Another example: $$\matrix{e&a&b\cr a&e&a\cr b&b&e\cr}\ ,\qquad (ab)a=e\ ,\qquad a(ba)=a\ .$$