Considering the set $S= \{-1, 1\}$, I need to show if associativity holds under binary operation multiplication. We used to take any three elements to verify the associative property, but here only two are given.
Can I consider any one twice ? like $1\cdot ((-1)\cdot (-1))= 1\cdot 1= 1$ and also $(1\cdot (-1))\cdot (-1)= (-1)\cdot (-1)= 1$, now in this case associativity holds, but I'm not sure if this is a valid approach, or should I say since the set has only two elements, we can't verify associativity.
The comments give useful approaches. Here are some supplementary notes and ideas. Throughout, assume $(S, *)$ is a set equipped with a binary operation.
Generally, there's Light's test for associativity: For each element $b$ of $S$ construct two binary operations on $S$: $$ \lambda_{b}(a, c) = (a*b)*c,\qquad \rho_{b}(a, c) = a*(b*c). $$ The notation is meant to suggest grouping on the left $\lambda$, and grouping on the right $\rho$. Associativity of $*$ is precisely that these operations are the same for all $b$. For small sets, constructing Cayley tables for these is feasible.
Further, if $(S, *)$ has an identity element $e$, the condition $(a * b) * c = a * (b * c)$ is automatic if any of the operands is $e$. (Right...?) In a set with two elements, consequently, there is really only one case to check!
Generally, the equality $\lambda_{e} = \rho_{e}$ in Light's test is automatic.