Here is another great Mark Sapir result. Let $S$ be the three element cyclic semigroup $\langle x\mid x^2=x^3\rangle$. Then $S$ has no finite basis for its quasi-identities. I believe Jackson and Volkov later showed that any finite semigroup containing this one also has no finite basis for its quasi-identities.

Related, Mark Sapir showed that although the variety generated by the finite semigroup $\{1,a,b\}$ where $1$ is the identity and $xy=x$ for $x,y\neq 1$ has only fnitely many subvarieties, it has uncountably many subquasivarieties. I recently showed with Margolis and Saliola that $\{1,a,b\}$ has no finite basis of quasi-identities (this is easier than Sapir's result) using hyperplane arrangements.