The result you quoted appears in this reference: G. Szasz, Die Unabhangigkeit der Assoziativitatsbedingungen, Acta. Sci. Math. Szeged 15 (1953), 20-28.

The Szasz theorem requires that the set $S$ have at least four elements, though it is also true for sets of size $3$.

Szasz' proof is constructive and goes as follows. Assuming $a$, $u$, $v$ and $w$ are distinct members of $S$, define $a\cdot a = u$, $a\cdot u = v$ and $x\cdot y = w$, for any $(x,y)$ other than $(a,a)$ and $(a,u)$. Then a case by case verification shows that $(a\cdot a)\cdot a = w\neq v = a\cdot(a\cdot a)$, but that every other triple $(x,y,z)$ associates.

Direct enumeration shows that there are (up to isomorphism) $10$ magmas of order $3$ with exactly one non-associative triple. (There are $124$ of order $4$.)

The result you quoted appears in this reference: G. Szasz, Die Unabhängigkeit der Assoziativitätsbedingungen, Acta. Sci. Math. Szeged 15 (1953), 20-28.

The Szasz theorem requires that the set $S$ have at least four elements, though it is also true for sets of size $3$.

Szasz' proof is constructive and goes as follows. Assuming $a$, $u$, $v$ and $w$ are distinct members of $S$, define $a\cdot a = u$, $a\cdot u = v$ and $x\cdot y = w$, for any $(x,y)$ other than $(a,a)$ and $(a,u)$. Then a case by case verification shows that $(a\cdot a)\cdot a = w\neq v = a\cdot(a\cdot a)$ but that, for every other triple $(x,y,z)\in S^3$, we have $(x\cdot y)\cdot z = x\cdot(y\cdot z)$.

Direct enumeration shows that there are (up to isomorphism) $10$ magmas of order $3$ with exactly one non-associative triple. (There are $124$ of order $4$.)