I Q sampling and baseband version of analytic signal
The problem in your derivation is the way you discard the high frequency components. Note that $\hat{s}(t)$, the Hilbert transform of the bandpass signal $s(t)$, has frequency components around $\omega_c$ as well as around $-\omega_c$. So you can't just discard one of the two terms $\hat{s}(t)e^{j\omega_ct}$ and $\hat{s}(t)e^{-j\omega_ct}$. The low pass filter filters out parts of both terms, but leaves other parts unchanged.

You could better represent $s(t)$ as

$$s(t)=\frac12\big[s_a(t)+s_a^*(t)\big]\tag{1}$$

where $s_a^*(t)$ is the complex conjugate of the analytic signal $s_a(t)$. This has the advantage that both components on the right-hand side of $(1)$ have only frequency components either at $\omega_c$ ($s_a(t)$) or at $-\omega_c$ ($s_a^*(t)$), but not at both frequencies.

So, for example, the in-phase component is easily derived as a low pass filtered version of $2s(t)\cos(\omega_ct)$:

$$\begin{align}I(t)&amp;=\textrm{LP}\big\{2s(t)\cos(\omega_ct)\big\}\\&amp;=\frac12\textrm{LP}\left\{\big[s_a(t)+s_a^*(t)\big]\left[e^{j\omega_ct}+e^{-j\omega_ct}\right]\right\}\\&amp;=\frac12\left[s_a(t)e^{-j\omega_ct}+s_a^*(t)e^{j\omega_ct}\right]\\&amp;=\textrm{Re}\left\{s_a(t)e^{-j\omega_ct}\right\}\tag{1}\end{align}$$

where $\textrm{LP}\{\cdot\}$ is the low pass filter operator.

In a similar way, the quadrature component $Q(t)$ can be shown to be a low pass filtered version of $-2s(t)\sin(\omega_ct)$.