Is there an easy way to see associativity or non-associativity from an operation's table? Most properties of a single binary operation can be easily read of from the operation's table. For example, given
$$\begin{array}{c|ccccc}
\cdot &amp; a &amp; b &amp; c &amp; d &amp; e\\\hline
a     &amp; e &amp; d &amp; b &amp; a &amp; c\\
b     &amp; d &amp; c &amp; e &amp; b &amp; a\\
c     &amp; b &amp; e &amp; a &amp; c &amp; d\\
d     &amp; a &amp; b &amp; c &amp; d &amp; e\\
e     &amp; c &amp; a &amp; d &amp; e &amp; b
\end{array}$$
it is easy to check that it is closed (no elements occur in the table which don't occur as row or column index), commutative (the table is symmetric), has a neutral element (the row and column of $d$ are copies of the index row/column), and has an inverse element for each element (there's a $d$ in each row and column). In other words, almost all important properties can immediately be seen. The only part missing is associativity.

Therefore my question: Is there a simple way to see directly from the operation's table (i.e. without doing explicitly all the calculations) if an operation is associative?
Have you seen Light's associativity test?  According to Wikipedia, "Direct verification of the associativity of a binary operation specified by a Cayley table is cumbersome and tedious. Light's associativity test greatly simplifies the task."

If nothing else, the existence of Light's algorithm seems to rule out the possibility that anyone knows an easy way to do it just by looking at the original Cayley table.

Note also that, in general, one cannot do better than the obvious method of just checking all $n^3$ identities of the form $(a\ast b)\ast c = a\ast (b\ast c)$.  This is because it is possible that the operation could be completely associative except for one bad triple $\langle a,b,c\rangle$. So any method that purports to do better than this must only be able to do so in limited circumstances.
A set with an operation used to build a four-dimensional multiplication table with entries $X_{abcd} = (A(BC))D$ is associativ iff $X_{abc1} = X_{1abc}$, with 1 being unity. So it would be a certain three-dimensional symmetry of a four-dimensional table.