A question from 1989 leningrad mathematical olympiad

Solution 1:

Note the associativity condition implies that parentheses are redundant, so something like $a \star b \star c \star d$ is uniquely-defined without parentheses.

Observe that, by axiom $3$, $1 = c \star d$ for some $c,d \in \mathbb{Z}$. A repeated application of axiom $3$ gives us that $d= e \star f$ for some integers $e,f$.

Hence we get that $1 = c \star e \star f$. Observe now that $-1 = f \star c \star e$ by the first axiom. A repeated application of the first axiom gives that $1 = e \star f \star c$. A third application gives $-1 = c \star e \star f$. But this is a contradiction, since we know that $1 = c \star e \star f$.