In a non-commutative monoid, is the left inverse of an element also the right inverse? Assume there is an element $a\in M$, where $M$ is a non-commutative monoid. If there exists $a b\in M$ such that $a * b = n$, where $n$ is the neutral, does it follow that $b * a = n$?

I want to say no, because the axioms which define a group shouldn't be redundant ($a * b = b * a = n$ is the 4th axiom). But the following logic seems to prove that the inverse is automatically associative:

$b * n = b$

$b * (a * b) = b$

$(b*a) * b = b$

Thus for the last line to be true, $b*a$ has to evaluate to $n$.
The answer is no. The bicyclic monoid is a counter-example. However, in a finite monoid, the answer is yes.
In your question, you called $M$ a non-commutative monoid. It should follow just from that, that the operations included in $M$ are not commutative. Thus, $a*b$ is not necessarily equal to $b*a$.

Normally, when someone twists my arm for an example of an associative but non-commutative binary operation, I answer "matrix multiplication."