totally ordered semigroups
Monogenic semigroups.
Consider the case of a [monogenic semigroup](monogenic semigroup), that is, a semigroup $S$ generated by a single element $s$. If $S$ is finite, then it can be totally ordered if and only if it is aperiodic, that is, if there exists a positive integer n such that $s^n = s^{n + 1}$. In this case, the order could be $s \leqslant s^2 \leqslant \dotsm \leqslant s^n$.
If $S$ is not aperiodic, then it contains a finite subgroup, and the only compatible order on a finite group is the equality, as you observed.

If $S$ is infinite, it can also be ordered by $s \leqslant s^2 \leqslant s^3 \leqslant\dotsm $.

Arbitrary semigroups. An interesting example is the free monoid $A^*$ equipped with some shortlex order. Shortlex orders are total and are compatible with the concatenation product.

Second question. I am not sure what you mean by a "finitely generated finite semigroup" since every finite semigroup is finitely generated. Thus I suppose that the question is whether the order of a finite ordered semigroup can be a total order. There are some examples which are not monogenic, for instance the ordered semigroup $(S, \leqslant)$ with $S = \{a,b\}$, $a \leqslant b$,  $aa = ba = a$ and $ab = bb = b$.

Third question. See my comment.