Does adjoining an identity to a cancellative semigroup that is not already a monoid always give a cancellative monoid?
The only way that this could fail is if a product in a cancellative semigroup that is not a monoid is equal to one of the factors. But in fact, this never happens, as the following proof shows:
Suppose that $a \cdot b = a$ in a cancellative semigroup $S$. Then, for any element $c$ of $S$, one has that $a \cdot b \cdot c = a \cdot c$, and then cancelling $a$ gives $b \cdot c = c$. Hence, $b$ is a left identity. Similarly, if $a \cdot b = b$, then for any $c \in S$, one has that $c \cdot a \cdot b = c \cdot b$, and cancelling $b$ gives $c \cdot a = c$, implying that $a$ is a right identity. In particular, any right identity is also a left identity and vice versa, so if $a \cdot b = a$ or $a \cdot b = b$, then $b$ (resp. $a$) must in fact be a two-sided identity, so $S$ must therefore be a monoid.