Question 1 either has a trivial answer or it is answered in my paper (with coauthors).  [REPRESENTATION THEORY OF FINITE SEMIGROUPS,
SEMIGROUP RADICALS AND FORMAL LANGUAGE THEORY][1] depending on what you require for a local ring.  Notice that a semigroup algebra is not usually unital.  Do local rings have to have identities for you? If so, then the only examples are groups and they are the ones you gave.  If you allow nonunital local rings, read on.

In any event, if $S$ is any semigroup, then $KS$ has the trivial homomorphism $KS\to K$ and so if it is to have a unique maximal left ideal, then the augmentation ideal (the kernel of this map) had better be it.  If $S$ is finite, then this is the same as the augmentation ideal being the radical, that is, nilpotent.


We prove in our paper  (Proposition 3.4) that if $S$ is a finite semigroup, then the augmentation ideal is nilpotent in characteristic $0$ if and only if $S$ is locally trivial, meaning $eSe=\{e\}$ for each idempotent $e$, and if the characteristic is $p>0$, then the augmentation ideal is nilpotent if and only if $S$ is locally a $p$-group, meaning $eSe$ is a $p$-group for each idempotent $e$.  In particular, the only time this can occur for a monoid is when it is a group, in which case you already knew the answer.  One can also deduce this from the much deeper Clifford-Munn-Ponizovskii theorem which tells you a semigroup algebra is local iff it has a unique maximal subgroup (up to $\mathcal J$-equivalence) and that maximal subgroup has a local group algebra.  But our proof avoids using deeper theory.

As to your second question, an algebra given by quivers with relations must be unital.  So in this case you are dealing either with the trivial group or a $p$-group in characteristic $p$ and I think we have discussed in the past on MO what quivers and relations come up this way. Semigroups will not give you new examples.

  [1]: https://www.ams.org/journals/tran/2009-361-03/S0002-9947-08-04712-0/S0002-9947-08-04712-0.pdf