Theorem. For every $\phi $ in $C_c(\hat G)$, one has that $\check \phi $ is a positive definite function if and only if $$ \phi (x)\geq 0, \quad\forall x\in \hat G. $$
Proof. I understand that the OP already knows how to prove the implication "$\Leftarrow$" (because then $\phi =|\phi |$), so I will only bother to prove "$\Rightarrow$".
We therefore pick $\phi $ in $C_c(\hat G)$, and assume that $\check \phi $ is a positive definite function. Denoting the duality of $G$ and $\hat G$ by $$ (t, x)\in \hat G\times G\mapsto \langle t, x\rangle \in {\mathbb C}, $$ we have for all $x_1, x_2, \ldots , x_n\in G$, and all $a_1, a_2, \ldots , a_n\in {\mathbb C}$, that $$ 0 \leq \sum_{i, j=1}^n \bar a_i a_j \check \phi (x_i^{-1}x_j) = \sum_{i, j=1}^n \bar a_i a_j \int_G\langle t, x_i^{-1}x_j\rangle \phi (t)\, dt = $$$$ = \int_G\sum_{i, j=1}^n \bar a_i a_j \overline{\langle t, x_i\rangle }\langle t, x_j\rangle \phi (t)\, dt = \int_G\Big |\sum_{j=1}^n a_j\langle t, x_j\rangle \Big |^2 \phi (t)\, dt =$$$$= \int_K|p(t)|^2\phi (t)\, dt, $$ where $K$ is the compact support of $\phi $, and $p(t)=\sum_{j=1}^n a_j\langle t, x_j\rangle $.
Observe that the collection of functions $p$ that one obtains by letting the $x_j$ range in $\hat G$, and the $a_j$ range in ${\mathbb C}$, gives a subalgebra of $C(K)$, containing the constant functions and separating the points of $K$. By Stone-Weierstrass, this algebra is uniformly dense in $C(K)$, whence the set of functions of the form $|p(\cdot)|^2$ is uniformly dense in the positive cone of $C(K)$.
The computation above then implies that $$ \int_K f(t) \phi (t)\, dt\geq 0, \qquad (*) $$ for every nonnegative continuous function on $K$.
We next claim that $\phi $ is a nonnegative function. To see this, assume by contradiction that there exists $t_0$ in $\hat G$ such that $\phi (t_0)$ lies outside the real interval $[0, +\infty )$. Since this interval is a closed subset of ${\mathbb C}$, there exists $\varepsilon >0$, such that the open ball $B_\varepsilon (\phi (t_0))$ has empty intersection with $[0, +\infty )$.
Set $$ U= \{t\in \hat G: |\phi (t)-\phi (t_0)|<\varepsilon /2\}, $$ so $U$ is an open subset of $\hat G$, contained in $K$.
Pick any nonnegative function $f$ in $C(K)$, vanishing on $K\setminus U$, and such $f(t_0)>0$. Since Haar measure has full support, we have that $\int_Kf(t)\, dt >0$, and if we multiply $f$ by a suitable positive number, we may assume that $$ \int_Kf(t)\, dt =1. $$
We then have $$ \Big |\phi (t_0) - \int_K f(t)\phi (t)\, dt\Big | = \Big |\int_K f(t)(\phi (t_0)-\phi (t))\, dt\Big | \leq $$$$ \leq \int_U f(t)|\phi (t_0)-\phi (t)|\, dt \leq \varepsilon /2<\varepsilon . $$ This shows that $$ \int_K f(t)\phi (t)\, dt \in B_\varepsilon (\phi (t_0))\cap [0, +\infty ), $$ a contradiction. QED
Using the Theorem it is now clear that $\check \phi $ is a constant multiple of a positive definite function if and only if $\phi $ is a constant multiple of a (pointwise) nonnegative function.
