I try to write down in full detail the answers of Nik and BS (many thanks to both!).
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)$.\\\