I have asked a question about exponentials yesterday, but there is something that is not clear to me so I thought I should ask a different question.
By looking at the case of sets, I am not sure about the following: If we suppose we have no idea of what the exponential of two sets $B$ and $C$ is, how does one identify it with $\text{Hom}(B,C)$? Why is the set $S$, endowed with a function $g:S\times B\to C$, such that for any set $A$ and function $f:A\times B\to C$ the property of the exponential holds (existence of a unique function that makes the triangle commute), the set $\text{Hom}(B,C)$ of functions from $B$ to $C$? When you know it, it is easy to show that it does indeed satisfy the definition, but if you don't?
By definition, the exponentiation of a set $S$, if it exists, is the right adjoint of the functor $ - \times S \colon \mathsf{Set} \to \mathsf{Set}, $ i.e. a functor $F_S \colon \mathsf{Set} \to \mathsf{Set}$ such that there is a bijection $$ \hom(A\times S, B) \simeq \hom(A,F_S(B)) $$ natural in the sets $A,B$.
As Zhen Lin pointed it out in the comments, take $A$ a singleton : one must have $$ \hom(\{\ast\} \times S, B) \simeq \hom(\{\ast\},F_S(B)), $$ which says that $F_S(B)$ (which is isomorphic to set on the right) must be isomorphic to $\hom (S, B)$ (which is isomorphic to the set on the left). So one try $F_S(-) := (-)^S$ and check that it indeed is a right adjoint to $- \times S$ (which is not hard, as you said).