Associativity means that for all $a, b $ and $c$ in $S$, $(a * b) * c = a * (b * c).$
Since $S$ has only two elements,
there are only two possibilities to check for $a$, two for $b$, and two for $c$;
altogether, there are $2\times2\times2=8$ possibilities to check:
$(0*0)*0=0*(0*0)$
$(0*0)*1=0*(0*1)$
$(0*1)*0=0*(1*0)$
$(0*1)*1=0*(1*1)$
$(1*0)*0=1*(0*0)$
$(1*0)*1=1*(0*1)$
$(1*1)*0=1*(1*0)$ and
$(1*1)*1=1*(1*1)$.
You listed the 6 relatively primes to $18$, and not the primes, which is correct.
Your Cayley table is also (somewhat) correct, but note that we are working in $\Bbb Z_{18}$, i.e. modulo $18$. That is, you should rather put the remainders of the calculated products modulo $18$.
For example, the last term is $17^2 =289\equiv 1\pmod{18}$ (because $288$ is even and its digits sum to $18$ which is divisible by $9$).
(By the way, if you keep on adding the digits in the numbers in your table until you reach 1 digit, then if the result is odd, it's just the remainder, if it's even, subtract $9$.)
But that's also a consequence of $17\equiv -1\pmod{18}$ (meaning that their difference is divisible by $18$), and that congruent numbers are interchangeable in modular arithmetic, just like equal numbers are so in normal arithmetic (i.e. one has e.g. $a\equiv b\pmod m\implies ac\equiv bc\pmod m$ and thus if also $c\equiv d$, then $ac\equiv bc\equiv bd$.)
So, here is a simplified version of the same Cayley table you wrote, but using smallest absolute value representatives:
$$\matrix{\times \\
& 1&5&7&-7&-5&-1\\
& 5&7&-1&1&-7&-5\\
& 7&-1&-5&5&1&-7\\
& -7&1&5&-5&-1&7\\
& -5&-7&1&-1&7&5\\
& -1&-5&-7&7&5&1}$$
You can also observe that $5$ generates this group: keeping multiplying by $5$ we receive the following cycle containing all group elements:
$$1\,\mapsto\, 5\,\mapsto \, 7\,\mapsto \, -1\,\mapsto \, -5\,\mapsto \, -7\,\mapsto \, 1\,\mapsto \, \dots $$
Best Answer
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 $u$:
$$ \begin{array}{l} \text{Example with identity u} \\ \begin{array}{c|ccc} \hline * & u & v & w \\ \hline u & u & v & w \\ v & v & w & u \\ w & w & u & v \\ \hline \end{array} \end{array} $$