Is $ H\left\{x_h(t)\right\} = H\left\{x\left(t-\frac{\pi}{2w_c}\right)\right\} $ where $H$ is Hilbert transform
It is quite unclear what you're actually asking, and most of your equations have no meaning. But one thing that you need to understand is that the equation

$$\mathcal{H}\{x(t)\}=x\left(t-t_0\right),\qquad t_0=\frac{\pi}{2\omega_0}=\frac{1}{4f_0}\tag{1}$$

only makes sense for sinusoidal functions $x(t)$ (of frequency $f_0$). For non-sinusoidal signals $x(t)$ the Hilbert transform of $x(t)$ can only be written as an integral:

$$\mathcal{H}\{x(t)\}=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{x(\tau)}{t-\tau}d\tau\tag{2}$$

where the improper integral is interpreted as the Cauchy principal value. In general, the Hilbert transform is not just a simple time shift.

For a sinusoidal signal $x(t)=\cos(2\pi f_0t+\phi)$ you can indeed show that the Hilbert transform corresponds to a delay of $t_0=1/(4f_0)$, i.e., in the frequency domain the following holds:

$$-j\,\textrm{sgn}(t)\,X(f)=X(f)e^{-j2\pi f t_0}=X(f)e^{-j\frac{\pi f}{2f_0}}\tag{3}$$