My question is primarily more about the convolution integral/theorem than the proof in question, but I wanted to give some idea of why I'm asking.
The Laplace transform of the convolution $$(f\star g)(t)=\int^t_0 f(\tau)g(t-\tau)\,\mathrm{d}\tau$$
produces the product of the Laplace Transform of $f$ and $g$, that is $\mathcal L \left[(f\star g)(t)\right]=F(s)G(s)$.
My question is -- is $(f\star g)(t)$ somehow equivalent to $(f\cdot g)(t)$? I heard this method is useful for proving that the transform is a homomorphism (specifically when proving that the transform of the product is the product of the transforms).