I try to write down in full detail the answers of Nik and BS (many thanks to both!). I do not accept this answer as definitive since I'm still interested in the last non-commutative part of my question.
The references in the proof are to Folland's book "A Course in Abstract Harmonic Analysis" and Rudin's book "Fourier Analysis on Groups".
Let $G$ be an LCA group, then {$\widehat\phi:\phi\in L^1(G)^+\cap \mathcal P(G) $}$=L^1(\widehat G)^+\cap \mathcal P(\widehat G)$.
Proof. Let us start proving the inclusion {$\widehat\phi:\phi\in L^1(G)^+\cap \mathcal P(G)$}$\subseteq L^1(\widehat G)^+\cap \mathcal P(\widehat G)$. Indeed, consider $\phi \in L^1(G)^+\cap \mathcal P(G)$, then $\widehat\phi\in L^1(\widehat G)$ by the Fourier Inversion Theorem (see [Rudin, page 22, line 1]) and $\widehat\phi\geq 0$ by [Folland, Corollary 4.23]. Let now $\mu_\phi$ be the non-negative and bounded (as $\phi\in L^1(G)^+$) regular measure defined on a generic Borel subset $E$ of $G$ by $\mu_\phi(E)=\int_{x\in E}\phi(x)d\mu(x)$ (here $\mu$ is a fixed Haar measure on $G$). One can show that $$\widehat\phi(\gamma)=\int_{x\in G}\phi(x)\gamma(-x)d\mu(x)=\int_{x\in G}\gamma(-x)d\mu_\phi(x)\, .$$ By Bochner's Theorem (see [Rudin, page 19]), $\widehat\phi\in \mathcal P(\widehat G)$.
On the other hand, let $\phi\in L^1(\widehat G)^+\cap \mathcal P(\widehat G)$ and $\psi$ be the function defined by $\psi(\gamma)=\phi(-\gamma)$ for all $\gamma\in \widehat G$. It is not difficult to see that $\psi\in L^1(\widehat G)^+\cap \mathcal P(\widehat G)$. By the first part of the proof, $\widehat\psi\in L^1( G)^+\cap \mathcal P( G)$ and, using Fourier Inversion Theorem (see [Folland, Theorem 4.32]), one obtains that $\widehat{\widehat\psi}=\phi$, which is therefore the Fourier transform of a function in $ L^1( G)^+\cap \mathcal P( G)$.\\\
