I will try (1) and go for infinite (the bonus).
The finite case is called the code problem: finding possible decompositions is like decoding. Its classical algorithm is called Sardinas–Patterson. It is not immediately clear it is a finite search space. The lengths of the decomposition might be long. Still it is not Post Correspondence as that has additional restrictions.
The finite case has a simple regular language solution: for every pair of distinct $s,t \in S$ test whether the regular set $s S^* \cap t S^*$ is empty. (The Sardinas-Patterson tries that "in parallel".) This does not work in the infinite case as there would be infinitely many $s,t$ to test.
In the infinite (but regular) case I would build a new finite state automaton based on an automaton for $S$. It simulates in parallel the two different decompositions by keeping track of a pair of states, one for each decomposition. On a letter both pairs make a step on the same letter. In a final state each component may either restart (it has seen an element in $S$) or proceed (trying to find a longer word in $S$). Now we need one additional boolean addition to the state-pair as we need to check whether the two components made different choices in at least one point. Accept if both simulations end in a final state (and we have seen a point where the two decompositions differed). Now again there are no two different decompositions for the same string iff the language of the automaton is empty.
And indeed, for context-free the problem must be undicidable, but I do not have a reference at hand now.