Are all function transforms special cases of Gelfand's transform?

Reading about Gelfand-Naimark theorem I've seen that the Fourier transform is a special case of Gelfand transform for the space $L^1(\mathbb{R})$ with the convolution product. In a related question on this site (Fourier transform as a Gelfand transform) I see (if I well understand) that this is a case of Pontryagin duality. Now my question is: also other transforms, as Laplace transform, are cases of Gelfand transform? And, if yes, what is the suitable algebra of functions in which we can see such correspondence?

If $L^1(\mathbb{R}^+)$ is equipped with the convolution product $$(f \star g)(t) = \int_0^tf(t − s)\cdot g(s) \,ds\qquad, \forall f, g \in L^1(\mathbb{R}^+)$$ then it becomes a Banach algebra. The Gelfand theory here corresponds to the Laplace transform.
• Thanks. You know if this result can be generalized to other transforms? As Hankel transform that use Bessel functions as a basis. And to functions on $\mathbb{R}^n$ ? Feb 25 '15 at 9:46