An exercise I'm working out suggests proving $1^a \cong 1$.
First of all, I've noticed that exponential objects seem to be unique up to isomorphism (even though I wasn't able to build a limit-like construction using the scarce knowledge of mine about limits and cones and all that, and my book doesn't explicitly mention this fact, so I had to prove this separately). In particular, let $e_1, e_2$ be two exponential objects for $a, b$, and let $\text{ev}_1, \text{ev}_2$ be the corresponding evaluation maps. Then consider this diagram, which obviously commutes:
Note that if we leave the middle $e_2 \times a$ out and consider the morphism $(\widehat{\text{ev}_2} \times \text{id}_a) \circ (\widehat{\text{ev}_1} \times \text{id}_a) = (\widehat{\text{ev}_2} \circ \widehat{\text{ev}_1}) \times id_a$, then the corresponding diagram will commute. But it also commutes if we consider the morphism $\text{id}_{e_1} \times \text{id}_a : e_1 \times a \rightarrow e_1 \times a$, which, by uniqueness, implies that $\widehat{\text{ev}_2} \circ \widehat{\text{ev}_1} = \text{id}_{e_1}$.
Swapping the $_1$ and $_2$ indices everywhere we get $\widehat{\text{ev}_1} \circ \widehat{\text{ev}_2} = \text{id}_{e_2}$.
Having proven that, let's turn back to our original question. Note that $1$ satisfies the universal property for the exponent $1^a$: let $f : c \times a \rightarrow 1$, and let $\text{ev} : 1 \times a \rightarrow 1$ (which exists because $1$ is terminal). Consider the morphism $\hat{f} \times \text{id}_a : c \times a \rightarrow 1 \times a$, where $\hat{f}$ is the unique morphism from $c$ into $1$. Now obviously $f = \text{ev} \circ \hat{f} \times \text{id}_a$ (because $f$ is unique by the universal property of $1$), so the diagram commutes, and we've proven that $1$ is the exponent of $1$ and $a$ (in the $1^a$ sense).
By the first part, this implies $1 \cong 1^a$, and we're done.
So, does this look good? Is there a better way? Is it possible to delegate the proof of uniqueness of exponential objects to some common limits construction?
Exponential objects, as any object defined by a universal property, are unique up to isomorphism. (This is a general fact: it could be helpful to prove this yourself, for example, by showing that adjoint functors are unique up to natural isomorphism.) To show that two objects are isomorphic, therefore, it suffices to show they satisfy the same universal property.
You've shown that both objects satisfy the universal property of being exponentials, and this is a perfectly fine way to prove the objects are isomorphic. However, I think it's slightly simpler to show they both satisfy the universal property of being terminal objects, as follows.
Let $\mathscr C$ be a cartesian closed category and let $A, B \in \mathscr C$. Consider the hom-set $\mathscr C(B, 1^A)$. By the definition of the exponential (as a right-adjoint), $\mathscr C(B, 1^A) \cong \mathscr C(B \times A, 1)$, which is a singleton by the universal property of the terminal object. Therefore $1^A$ is itself terminal, as there is a unique morphism to it from any object $B$.
Finally, note that exponential objects are not limits or colimits (at least not in the category $\mathscr C$).