Is the Laplace transform a functor?

I believe the answer is "yes", but not between differential equations and algebraic equations. First, let's recall some facts about the Fourier transform. The Fourier transform can be defined on any locally compact abelian group and the category of all such groups is usually denoted $\mathbf{LCA}$. On any such group there exists a unique translation invariant measure called the Haar measure which allows all the usual definitions to carry over to this more general setting. There's a collection of results called Pontryagin duality which answers your question for the Fourier transform (it is a functor) and also generalizes familiar results about the Fourier transform on the usual spaces $\mathbb{R}, \mathbb{T}, \mathbb{Z}$ and $\mathbb{Z}/n\mathbb{Z}$ to any LCA group.

What does it mean to say that the Fourier transform is a functor? To any LCA group $G$ can identify its dual object $\hat{G}$, the collection of all continuous homomorphisms $\chi : G \rightarrow \mathbb{T}$ called characters. For instance, the dual of $\mathbb{T}$ is $\mathbb{Z}$ and this gives the theory of Fourier series. It turns out that $\hat{G}$ is also an LCA group. Now suppose we have two groups $G$ and $H$ in $\mathbf{LCA}$ and a homomorphism $f : G \rightarrow H$. There is a canonical way to come up with another homomorphism $\hat{f} : \hat{H} \rightarrow \hat{G}$. If $\chi \in \hat{H}$ and $g \in G$, then $$\hat{f}(\chi)(g) = \chi(f(g)).$$ The process of assigning to each group in $\mathbf{LCA}$ its dual group is functorial.

Since the dual of an LCA group is also an LCA group, we can ask about the double dual. And Pontryagin duality says that the map $G \rightarrow \hat{\hat{G}}$ is a really nice one. This is analogous to the double dual of a finite dimensional vector space.

So what about the Laplace transform? Now to each LCA group $G$ we look at the continuous homomorphisms $\chi : G \rightarrow \mathbb{C}^{\times}$. We have changed the codomain from the unit circle to the multiplicative group of complex numbers. It turns out that the collection of all such homomorphisms is the topological group $\text{Hom}(G,\mathbb{C}^{\times})\simeq\hat{G} \times \text{Hom}(G,\mathbb{R})$. For a reference for all this, see the book Commutative Harmonic Analysis or this paper by Mackey.

As for whether or not the Laplace transform is a functor, you play the same game as before. Let's take two groups $G$ and $H$ in $\mathbf{LCA}$ and a continuous homomorphism $f : G \rightarrow H$. Is there a canonical way to define a map $$\hat{f} : \hat{H}\times\text{Hom}(H,\mathbb{R}) \rightarrow \hat{G}\times\text{Hom}(G,\mathbb{R})$$ so that the assignment acts as a functor? Usually there's Only One Way to do these things, so let's try. Given an element $(\chi,\lambda) \in \hat{H}\times\text{Hom}(H,\mathbb{R})$ and $g \in G$ we want to define a map $\hat{f}(\chi,\lambda)$ which eats $g$ and spits out an element of $\mathbb{C}^{\times}$. Consider $$\hat{f}(\chi,\lambda)(g) = \chi(f(g))e^{\lambda(f(g))}$$ where I have used the notation of the booked linked above.

You should check whether this process satisfies the properties of a functor. I'm fairly certain it does, but am not that knowledgeable in category theory. You'd also have to ask, what is the target category? I think it would be the collection of all topological groups of the form $\hat{G}\times\text{Hom}(G,\mathbb{R})$ for some $G \in \mathbf{LCA}$ and the restricted set of morphisms induced by the continuous homomorphisms $G \rightarrow H$. Maybe someone with more categorical knowledge will come along and comment on this answer, which is too long for a comment?