Extending the definition of convolution product on $\mathcal{F}(C(G))$, $G$ abelian compact group

Question

Let $G$ be a compact abelian group. We know that we can define the convolution product on its convolution algebra $L^1(G)$ in a natural way: \begin{equation} (f_1*f_2)(g):=\int\limits_Gf_1(h)f_2(h^{-1}g)\mathrm{d}h \end{equation} ($L^1(G)$ has a structure of abelian Banach $^*$-algebra when endowed with this product, a proper involution and a natural norm). Now, the Riemann-Lebesgue theorem holds in this abstract setting too, therefore we can write the Fourier transform on $L^1(G)$ as a map \begin{aligned} \mathcal{F}\colon\,\,(L^1(G),\|\cdot\|_1)&\to(C_0(\widehat{G}),\|\cdot\|_\infty)\\ f&\mapsto\left[\omega\mapsto\int\limits_{G}f(g)\overline{\omega(g)}\mathrm{d}g\right] \end{aligned} This map is an injective homomorphism of Banach $^*$-algebras (actually, $C_0(\widehat{G})$ is even a $C^*$-algebra and its Pontryagin dual $\widehat{G}$ is discrete). Now, since $G$ is compact, $C(G)\subset L^1(G)$ hence the Fourier transform is surely well defined on $C(G)$. I am wondering if it possible to define the convolution product on $\mathcal{F}(C(G))\subset C_0(\widehat{G})$ as \begin{equation} \mathcal{F}(f_1)*\mathcal{F}(f_2):=\mathcal{F}(f_1f_2) \end{equation} for every $f_1,f_2\in C(G)$. On $L^1(\widehat{G})\subset C_0(\widehat{G})$ this definition surely works: it simply says that the Fourier antitransform respects the products, which is true. I searched for clues in many sources but I did not find anything useful. If anyone knows references about it in abstract harmonic analysis, that's well appreciated.