Let $\{Im (s)\lt 0\}=\{s\in \mathbb{C}\mid Im(s)\lt 0\}$, and $H^2(\{Im (s)\lt0\})$ is the Hardy space on $\{Im (s)\lt 0\}$. I know a classical theorem of Paley and Wiener
Fourier transform $\mathscr{F}$ is an unitary operator from $L^2(0,\infty)$ to $H^2(\{Im (s)\lt0\})$.
How can I conclude that Mellin transform is an unitariy operator from $L^2[0,1]$ to $H^2(\{Re (s)\gt \frac{1}{2}\})$ from the above Paley-Wiener theorem?
This has nothing to do with complex analysis or Paley-Wiener theorems; it's just the Plancherel theorem plus a change of variables. Leaving out the $2\pi$s:
Say $\int_0^\infty |f(x)|^2\,dx<\infty$. Let $$g(t)=f(e^{t})e^{t/2}.$$Then $$\int_{-\infty}^\infty|g(t)|^2\,dt=\int_0^\infty|f(x)|^2\,dx$$and $$Mf(s)=\int_0^\infty f(x)x^{-1/2-is}\,dx=\hat g(s).$$So $$||Mf||_{L^2(\Bbb R)}=||g||_{L^2(\Bbb R)}=||f||_{L^2((0,\infty))}.$$