Laplace Transform in the context of Gelfand/Pontryagin

Question: Do quasi-characters or some other semi-group properly generalize the Laplace transform or decompose functions in some setting in a way similar to how characters generalize the classical Fourier transform and decompose $L^1$ functions for locally compact abelian groups?

For all locally compact commutative groups $G$ with Haar measure $\mu$ and character group $\widehat{G}$, the Fourier transform

\begin{equation} \mathcal{F}(f)(\chi)= \int_G f(x)\overline{\chi(x)} d\mu(x) \end{equation}

takes $L^1(G,\mu) \to \text{C}_\infty(\widehat{G})$ (the decaying continuous functions on $G$). Under convolution, $L^1(G,\mu)$ is a Banach algebra, so it makes sense to talk about the Gelfand transform. If $\phi \in \mathcal{M}_B$ (the maximal ideal space, or space of nonzero characters on the algebra $L^1(G,\mu)$) then $\phi$ is also a member of the unit ball in the dual space $B^*$ (which is where we get the topology for $\mathcal{M}_B$), so we can say that there is some $\alpha_\phi \in L^\infty(G,\mu)$ such that

\begin{equation} \widehat{f}(\phi) = \phi(f) = \int_G f(x) \overline{\alpha_\phi(x)}d\mu(x) \end{equation}

Using the convolution structure of $L^1(G,\mu)$, it an be proven that $\alpha_\phi = \chi$ for some character $\chi \in \widehat{G}$, and so the Gelfand transform and Fourier transform actually coincide here (we essentially have $\mathcal{M}_B \cong \widehat{G}$). Also when talking about $L^1(G,\mu)$, the Gelfand transform is injective.

We started with characters $\chi\in \text{Hom}(G,\mathbb{T})$, but we could also consider $\text{Hom}(G,\mathbb{C}^\times)\cong \widehat{G} \times \text{Hom}(G,\mathbb{R})$. These are sometimes called generalized or quasi-characters, and the $\text{Hom}(G,\mathbb{R})$ are called real characters. Quasi-characters for $\mathbb{R}$ look like $\chi(x) = e^{(\sigma + it) x}$, so a Fourier transform for quasi-characters is the Laplace transform

\begin{equation} \widetilde{f}(\chi) = \int_{-\infty}^\infty f(x) e^{(\sigma - it)x}dx \end{equation}

Unlike the regular Fourier transform, this splits the quasi-characters into semigroups $\sigma \le 0$ and $\sigma \ge 0$, which are defined on the separate components of $L^1(\mathbb{R}) = L^1(\mathbb{R}_+)\oplus L^1(\mathbb{R}_-)$. Similarly, the quasi-characters on $\mathbb{Z}$ look like $\chi(n) = z^n$ for $z \in \mathbb{C}^\times$, so that the Fourier transform for quasi-characters is a Laurent series

\begin{equation} \widetilde{f}(\chi) = \sum_{n=-\infty}^\infty f(n)z^n \end{equation}

which splits the quasi-characters again into semigroups $|z|\le 1$ and $|z|\ge 1$ for the separate components $\ell^1(\mathbb{Z}) = \ell^1(\mathbb{N}_0)\oplus \ell^1(\mathbb{N}_0^-)$.

Both of the scenarios lead to Hardy spaces, but this construction is specific to $\mathbb{R}$ and $\mathbb{Z}$. Is there a general Hardy space construction for semigroups of this kind? Compact commutative groups, for instance, have no real characters since it would map them into a compact subgroup of $\mathbb{R}$ (of which there is only $\{0\}$), and so a generalization would necessarily require a different kind of semi-group constructed from $G$.

Pontryagin duality relies on the fact that the irreducible unitary representations of commutative groups are 1-dimensional, so the characters (which act trivially on such representations) have a natural group structure. Similar to how the Gelfand transform "decomposes" $\text{C}_\infty(X)$ functions in terms of their values at points of $X$, a commutative group structure creates a natural way to decompose $L^1$ functions (for which point evaluations don't make sense) by their points in a "frequency" space. I think of this as a similar phenomena to the Gelfand transform on an $L^\infty$ space, which creates a new topology for the highly discontinuous $L^\infty$ functions to act continuously on, or the transform on $\text{C}_b(X)$ compactifying the space. In this vein, I've heard people say that the Laplace transform decomposes functions in terms of an "energy" space, can this be made precise at all?