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 $f(t)=\text{constant}$.
For the case of general $b$ I proceed as follows; substitute $f(t)$ by $$f(t)=e^{\frac{1}{2} (1-b)^2 \omega^2} e^{(1-b) b \omega t} e^{\frac{1}{2} (bt)^2}h(t).$$ Then one has $$g_\omega(t)\equiv e^{-(bt+\omega)^2/2}f(t+\omega)=h(t+\omega).$$ So we are back to case we studied earlier, and we can conclude that $h(t)=\text{constant}$. We thus arrive at the general solution $$f(t)=\text{constant}\times e^{(1-b) b \omega t} e^{\frac{1}{2} (bt)^2}.$$ Note that the earlier special cases $b=0$ and $b=1$ are recovered.