Let's pretend the Fourier transform was never invented.
Well, how could we prove there exists a linear transform $L$ over complex numbers such that, for two convolved integrable functions $f*g$, that $L[(f*g)(t)] = k \cdot L(f)L(g)$ for some normalization constant $k$?
I don't know much about abstract algebra, but this almost looks like the definition I see of a homomorphism. Are there circumstances under which we can ignore $k$?
I don't know the precise answer off the top of my head and it quickly gets quite advanced but there is basically a whole theory of this in abstract harmonic analysis. The Fourier transform is a type of Gelfand representation.
https://en.wikipedia.org/wiki/Gelfand_representation#Examples
We start with the algebra $A = L^1(\mathbf{C})$, where the algebra multiplication is convolution and let $\Phi$ denote the space of all multiplicative linear functionals on $A$, i.e. algebra homomorphisms $\phi : L^1(\mathbf{C}) \to \mathbf{C}$. Then for $f \in L^1(\mathbf{C})$ we can define $\tilde{f} : \Phi \to \mathbf{C}$ by $$ \tilde{f}(\phi) = \phi(f).$$
So the association $f \mapsto \tilde{f}$ defines a map $L^1(G) \to $ {some space of functions on $\Phi$}.
So I guess (now someone will have to correct me if I'm wrong), the point is that you can show this is an algebra homomorphism to $C_0(\Phi)$ and that actually $\Phi \simeq \mathbf{C}$.