I'm not quite sure I understand your question, but some comments. The intuition that exponentials generalize products really only makes sense in categories that resemble $\text{Set}$ and in general exponentials can look quite different from this. A nice example to think about is Heyting algebras, where the exponential generalizes implication in propositional logic (to intuitionistic logic). Here eval becomes modus ponens.
In any case, once you have composition you have eval provided that you believe a very small thing. Let me write $A^B$ as $[B, A]$. The composition map $[A, B] \times [B, C] \to [A, C]$ specializes, when $A = 1$, to a map
$$[1, B] \times [B, C] \to [1, C]$$
which reproduces eval as soon as you believe that there should be a natural isomorphism $[1, A] \cong A$.
Edit: Okay, so I think I understand your question better now. In fact it is not necessary to relate exponentials in a suitably generalized sense to products if you don't want to. You can write down the axioms of a closed category instead, although I don't know any natural examples which are not closed monoidal categories. In a closed monoidal category the cartesian product is replaced by another (usually symmetric) monoidal structure; the prototypical example is $\text{Vect}$ equipped with the tensor product.
The answer you probably don't want is that these kinds of "implementation details" really don't matter at all. Pick whichever makes you happiest, and it will never impact the answers to category-theoretic questions.
First, let me give a simpler example to make sure that I understand what you're asking:
We have "two possible implementations" for a group homomorphism $(G, \cdot) \to (H, \star)$.
- A triple $((G,\cdot),(H,\star),f)$ where $f : G \to H$ is a map on underlying sets (satisfying certain conditions)
- A function $f : G \to H$ on the underlying sets (satisfying certain conditions)
Now you worry that, under definition $2$, two different homomorphisms can have the same implementation. For instance, if we define $\cdot^\text{op} : G \times G \to G$ as $a \cdot^\text{op} b = b \cdot a$, then the same function $f$ can represent both a homomorphism $(G, \cdot) \to (H, \star)$ and to $(G,\cdot^\text{op}) \to (H,\star^\text{op})$ because these two homomorphisms have the same underlying function!
In definition $1$, we explicitly "tag" our function with its domain and codomain as groups so that the two homomorphisms $((G,\cdot), \ (H,\star), \ f)$ and $((G,\cdot^\text{op}), \ (H, \star^\text{op}), \ f)$ are encoded by different objects in the ambient set theory.
If I understand correctly, you're asking basically the same question, but in the context of categories and functors rather than groups and homomorphisms.
Now, in the group theoretic case and the category theoretic case, these implementation details don't matter. You will never be presented with a naked function out in the wild (or pair of functions in the category case) and asked "what is the group structure on the domain of this function?". Ironically, given the question, we can make this precise using category theory! Whenever we interact with a homomorphism $f$, we're really interacting with $f \in \text{Hom}((G,\cdot), (H,\star))$! We always have the bonus context on hand of what the domain and codomain should be. So, if you want${}^1$, you can absolutely use definition $1$ and tag your homomorphisms (functors) with the group structure (category structure) on their source and target.
Why does category theory make this precise? Because we can define categories of groups $\mathsf{Grp}_1$ and $\mathsf{Grp}_2$ using the two implementations of homsets. It's an easy exercise to show that these categories are equivalent (in fact, isomorphic!) so that, in this precise sense, you cannot distinguish the two definitions using only group theoretic questions. Similarly, you can build two categories $\mathsf{Cat}_1$ and $\mathsf{Cat}_2$ using the two definitions of functor, and you can show that these categories are also isomorphic${}^2$! So the answer to any "category theoretic question" you can ask won't depend on whether you choose definition 1 or 2!
I hope this helps ^_^
${}^1$: and it seems like you do want this, since you say you're doing this for your arrows in $\mathsf{Set}$
${}^2$: you do have to eat your tail a little bit since checking isomorphism of categories requires working in a category of categories. And in this category of categories you need a notion of functor, so you have to choose a definition before you can prove the two definitions agree... But this is really no different than any other time we have two equivalent definitions and we have to choose one "official" one to write down first.
Best Answer
After pondering about Zhen Lin's comment, it seems to me that this only a theoretical problem.
If we only know that $\mathcal C$ has "things" described via a universal property, then we get indeed a choice problem. Since the objects of $\mathcal C$ in general form a proper class, we would need a version of the axiom of choice for classes. My knowledge about set theory is limited, but I have never seen such a general choice axiom and can't judge which problems it could cause.
The general pattern of a universal property is this: We are given a diagram $\Delta$ in $\mathcal C$ consisting of a family of objects $(X_i)_{i \in I}$ and a family of morphims between these $X_i$. Then we consider diagram extensions $\Delta^*$ obtained from $\Delta$ by adding a single object $X(\Delta^*)$ and morphisms between $X(\Delta^*)$ and the $X_i$ such that some characteristic condition $\mathfrak P$ is satisfied. We consider two types of diagram extensions: Sink extensions in which all added morphisms have codomain $X(\Delta^*)$, i.e. have the form $f_i^{\Delta^*} : X_i \to X(\Delta^*)$, and source extensions in which all added morphisms have domain $X(\Delta^*)$, i.e. have the form $f_i^{\Delta^*} : X(\Delta^*) \to X_i$.
Let us focus on sink extensions; everything can of course be formulated "dually" for source extensions.
A morphism from a sink extension $\Delta_1^*$ to a sink extension $\Delta_2^*$ is a morphism $\phi: X(\Delta_1^*) \to X(\Delta_2^*)$ such that $f_i^{\Delta_2^*} \circ \phi = f_i^{\Delta_1^*}$ for all $i \in I$.
A sink extension $\Delta_u^*$ is called a universal sink extension of characteristic $\mathfrak P$ if for each source extension $\Delta^*$ there exists a unique morphism $\phi : X(\Delta^*) \to X(\Delta_u^*)$.
Equalizers, kernels, products and limits are examples of universal sink extensions; coequalizers, cokernels, coproducts and colimits are examples of universal source extensions. So perhaps we should call a source extension a cosink extension.
How do we know that a diagram $\Delta$ in a category $\mathcal C$ has a universal sink or source extension?
Certainly we have to give a proof, and this requires an explicit construction assigning to $\Delta$ an object $X_u(\Delta)$ and morphisms between $X_u(\Delta)$ and the $X_i$. Such a constructive existence proof usually will not involve any choices; it works like a (deterministic) algorithm.
In other words, we get a function assigning to an input consisting of a diagram $\Delta$ an output consisting of a specific universal extension $\Delta^*_u$.
Therefore, if we have some category $\mathcal C^*$ of diagrams in $\mathcal C$ with universal universal sink or source extensions of characteristic $\mathfrak P$, we get a functor $$\mathfrak P^* : \mathcal C^* \to \mathcal C^s$$ where $\mathcal C^s$ denotes the category of sink or source diagrams in $\mathcal C$.