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 $f(t)=\text{constant}$.

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For the case of general $b$ I could 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. Also check that
$$\int_{-\infty}^\infty e^{-(bt+\omega)^2/2} f(t+\omega) \frac{2}{i  t} dt=\text{constant}\times\int_{-\infty}^\infty dt/t=0,$$
if I interpret the singular integral in the OP as $\lim_{b\rightarrow\infty}\lim_{a\rightarrow 0}\left(\int_{-b}^{-a}dt/t+\int_{a}^b dt/t\right)$.

Since this solution is $\omega$-dependent it is not a useful answer. I am inclined to think there is no $\omega$-independent solution for $b\neq 0,1$.
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