Given a Banach space with a multiplication (usually convolution in $L^p$-spaces) such that $\|x\cdot y\|\leq C \|x\|\cdot \|y\|$ we do have a Gelfand transform which on $L^1(\mathbb{R})$ coincides with the Fourier transform or in the case of $L^1(\mathbb{T})$ the Fourier coefficients (here $\mathbb{T}=\mathbb{R}/\mathbb{Z}\sim(-\pi,\pi]$ is the circle group).
That $\Omega$ is bounded is not really sufficient in order to have a worthy Fourier transform. Surely, for bounded $\Omega$ and $f\in L^p(\Omega)$ we may define $\hat{f}(\xi)=\int_\Omega f(x)e^{-ix\xi}dx$ Now, by defining $f=0$ on $\mathbb{R}\setminus\Omega$ we extend $f$ to $L^p(\mathbb{R})$, and then we may consider a translate of $f$, i.e. $f_a(x)=f(x-a)$, for which we have $\int_\mathbb{R} f_a(x)e^{-ix\xi}dx=\int_\mathbb{R} f(x-a)e^{-ix\xi}dx=\\\int_\mathbb{R} f(x)e^{-i(x+a)\xi}dx=e^{-ia\xi}\int_\mathbb{R} f(x)e^{-ix\xi}dx=e^{-ia\xi}\hat{f}(\xi)$ This is a problem because if $f$ is nice enough, we would have an inverse Fourier transform and the translate must then belong to $L^p(\Omega)$ - but that is not the case unless we have some additional algebraic structure on $\Omega$.