It may be helpful to rewrite this in a way that avoids the principal value:
$$0=\int_{-\infty}^\infty e^{-(bt+\omega)^2/2} f(t+\omega) \frac{2}{i  t} dt=\int_{-\infty}^\infty dt \int_{-\infty}^{\infty} dx\,e^{-(bt+\omega)^2/2} f(t+\omega) \,\text{sign}(x) e^{-ixt},$$
then define $g_\omega(t)=e^{-(bt+\omega)^2/2}f(t+\omega)$ with Fourier transform $G_\omega(x)=\int_{-\infty}^\infty g_\omega(t)e^{-ixt}\,dt$, and arrive at
$$0=\int_{-\infty}^\infty dx\,G_\omega(x)\,\text{sign}(x),\;\;\forall\omega\in\mathbb{R}.$$

For $b=1$ we have the identity $g_\omega(t)=g_0(t+\omega)$, hence $G_\omega(x)=e^{i\omega x}G_0(x)$. The only solution is then that $G_0(x)=\text{constant}\times\delta(x)$, hence $f(t)=\text{constant}\times e^{t^2/2}$. Similarly, for $b=0$ the only solution is $f(t)=\text{constant}$.

I do not know how to solve the case of general $b$.