We consider a small category $\mathcal{C}$ with binary products, and we consider, for any objects $A, B$ of $\mathcal{C}$, the assignments
\begin{eqnarray*} F :\,\, &\mathcal{C}^\text{op} &\longrightarrow \textbf{Set} \\ &X &\mapsto \text{Hom}_{\mathcal{C}}(X \times A, B) \\ &X'\xrightarrow{\bar{f}} X &\mapsto - \circ f\times \text{id}_A. \end{eqnarray*}
After some fidgeting with compositions and reversals of the order of morphisms in opposite categories, I managed to prove this is a functor, and therefore, in fact, arbitrariness of $A, B$ will show that the category is Cartesian closed (a ccc) if $B^A$ exists with a suitable evaluation map.
I want to be constructive about this, and to avoid confusion (for myself, mainly), I will be very formal because my syllabus is, and I don't want to diverge from it. Their, its definition is given via a universal property I will write down explicitly.
Datum: $F$ is a representable presheaf.
Required to prove: $(\exists E \in \mathcal{C}_0)(\exists e : E \times A \to B)(\forall D' \in \mathcal{C}_0)(\forall h \in \text{Hom}_\mathcal{C}(D'\times A, B))(\exists ! H \in \text{Hom}_\mathcal{C}(D', D))(h = e \circ H \times \text{id}_A)$
As we assume the $F$ is representable-- $F \cong \text{Hom}_\mathcal{C}(-, D)$ for some object $D$ -- our first candidate for $E$ will be this $D$. We can now simplify the demonstrandum slightly:
R.T.P.: $ (\exists e : D \times A \to B)(\forall D' \in \mathcal{C}_0)(\forall h \in \text{Hom}_\mathcal{C}(D'\times A, B))(\exists ! H \in \text{Hom}_\mathcal{C}(D', D))(h = e \circ H \times \text{id}_A)$.
Note that there is a natural isomorphism $\mu : F \,\tilde \Longrightarrow \,y(D)$. Consider its component $\mu_D : \text{Hom}_{\mathcal{C}}(D \times A, B) \, \tilde\longrightarrow \text{Hom}_{\mathcal{C}}(D, D) \ni \text{id}_D$ at $D$, and define $\delta := \mu_D^{-1}(\text{id}_D) : D \times A \to B$. This will be our canditate for $e$. Now we let $D' \in \mathcal{C}_0$ and $h \in \text{Hom}_\mathcal{C}(D'\times A, B)$ be arbitrarily given. Our goal is now to provide (existence of) a unique $H: D' \to D$ such that $h = \delta \circ H \times \text{id}_A$.
As $\mu$ is a natural transformation, the naturality square gives, for any $f : D' \to D$, the equality $$\mu_D(\delta) \circ f = (- \circ f) \circ \mu_D (\delta) = y(D)(f) \circ \mu_D (\delta) = \mu_{D'} \circ F_1(f)(\delta) = \mu_{D'} \circ (- \circ f \times \text{id}_A)(\delta) = \mu_{D'}(\delta \circ f \times \text{id}_A). $$
However, by construction, $$f = \mu_D(\mu_D^{-1}(\text{id}_D)) \circ f = \mu_D(\delta) \circ f = \mu_{D'}(\delta \circ f \times \text{id}_A). $$
I only now realise, that if there is any $H : D' \to D$ which works, it must be $H := f$, as $f$ is now completely determined. (Right..? No, wait, not right, because $f$ is still defined in terms of itself: $f = \mu_{D'}(\delta \circ f \times \text{id}_A)$. Confusing..)
Regardless, we can now simply compute the expression $\delta \circ H \times \text{id}_A$, of which we want it to be equal to our arbitrary $h : D' \times A \to B$. Here we go: $$\delta \circ H \times \text{id}_A = \mu_D^{-1}(\text{id}_D) \circ (\mu_{D'}(\delta \circ H \times \text{id}_A) \times \text{id}_A)). $$
Everything typechecks: $\mu_{D'}(\delta \circ H \times \text{id}_A) : D' \to D$, so $\mu_{D'}(\delta \circ H \times \text{id}_A) \times \text{id}_A : D' \times A \to D \times A$. As $\mu_D^{-1}(\text{id}_D)) \in \text{Hom}_\mathcal{C}(D \times A, B)$, the two perfectly compose to a map $D' \times A \to B$, That's good news, as $h$ also has these domain and codomain.
Unfortunately, I have no idea why they should in fact be equal! If feels like I'm 95% of the way there, but this is where I get stuck. If so, can anyone give me the final push? If not, where did I go wrong?
Use the naturality of the given isomorphism $\mu : \mathcal{C}(- \times A, B) \cong \mathcal{C}(-,E)$.
As you have observed, the arrow $e : E \times A \rightarrow B$ is just $(\mu_E)^{-1}(1_E)$.
Now, take any morphism $C\times A \xrightarrow{h} B$ and as you said, take $H := \mu_C(h)$.
The naturality of the above isomorphism applied to the arrow $C \xrightarrow{H} E$ gives you a commutative square :
$\require{AMScd}$ \begin{CD} \mathcal{C}(E \times A, B) @>{\mu_E}>> \mathcal{C}(E,E)\\ @V{- \circ (H \times 1_A)}VV @V{- \circ H}VV\\ \mathcal{C}(C \times A, B) @>{\mu_C}>> \mathcal{C}(C, E) \end{CD}
Applying the two (equal) composites in this diagram on $e$ and using the fact that the component of $\mu$ at $C$ is an isomorphism gives you $h = e \circ (H \times 1_A)$.
Now, suppose that $C \xrightarrow{H'} E$ is another arrow such that $h = e \circ (H' \times 1_A)$.
Again, using naturality, you get a similar commutative square :
$\require{AMScd}$ \begin{CD} \mathcal{C}(E \times A, B) @>{\mu_E}>> \mathcal{C}(E,E)\\ @V{- \circ (H' \times 1_A)}VV @V{- \circ H'}VV\\ \mathcal{C}(C \times A, B) @>{\mu_C}>> \mathcal{C}(C, E) \end{CD}
Applying the two equal composites again on $e$ says that $\mu_C(h) = H'$ which gives $H = H'$.