Prove that the operation in the following Cayley table has no identity element:
$$ \begin{array}{c|ccc} \hline * & u & v & w \\ \hline u & u & w & w \\ v & v & v & v \\ w & w & u & v \\ \hline \end{array} $$ The only way I can think is just by checking it one by one. But I don't think that be considered a proof.
Since there are only three elements, you can certainly "just" check each case. Indeed, that's exactly what you should probably do: "Proof by cases" is a perfectly legitimate method of proof:
Consider each element as a separate "case" (there are only three to consider). For each element $x \in \{u, v, w\}$, find a counterexample which shows that $x$ cannot be the identity.
For example, suppose we test the case where $x = w$: From the table, we have that $w*w = v.\;$ This means $w$ cannot be an identity, since if $x = w$ were the identity, we must have $w \times w = w$. $\;\checkmark$
Do the same for each of $u$ and $v$, and you're done.
One observation, as noted below in the comments: See if you can prove that if an identity element $x$ exists, we would need to have one column for element $x$ replicate the left-most column, and the corresponding row for $x$ replicate the top most row, the "header" row.
Example where there exists an identity element $u$:
$$ \begin{array}{l} \text{Example with identity u} \\ \begin{array}{c|ccc} \hline * & u & v & w \\ \hline u & u & v & w \\ v & v & w & u \\ w & w & u & v \\ \hline \end{array} \end{array} $$