The integral in question can be written as a Fourier transform, that is,
$$\mathcal{F}_{\omega \to s} \left\{ \frac{\sigma_0}{2\pi} \frac{1}{1-i\tau \omega}\right\}(s) = \int_{-\infty}^\infty \frac{d\omega}{2\pi} \frac{\sigma_0}{1-i\tau \omega}e^{-is\omega}$$
where $s \equiv t-t'$. This has a pole at $\omega = -i/\tau$, and from the Laurent series,
$$f(\omega)= \frac{\sigma_0}{2\pi} \left(\frac{i}{\tau}e^{-s/\tau}(\omega + i/\tau)^{-1} + \frac{s}{\tau}e^{-s/\tau} + \mathcal{O}((\omega+i/\tau)^2) \right)$$
one finds the residue is,
$$\mathrm{Res}(f,-i/\tau) = \frac{i\sigma_0}{2\pi\tau}e^{-s/\tau}.$$
Now consider the same integral, over the original real line contour $C_1$, plus an additional contour $C_2$ which lies in the complex plane, enclosing the pole to form a Jordan curve. We then have,
$$\int_{C_1 + C_2} \frac{d\omega}{2\pi} \frac{\sigma_0}{1-i\tau \omega}e^{-is\omega} = 2\pi i \, \mathrm{Res}(f,-i/\tau) = -\frac{\sigma_0}{\tau}e^{-(t-t')/\tau}.$$
We choose $C_2$ as the circular contour in the lower half plane, with a radius $R$, parametrised as,
$$\omega(\theta) = Re^{-i\theta}, \quad \theta \in [0,\pi].$$
Now, we seek a bound using the estimation lemma for the contour integral, given by,
$$\left| \int_{C_2} d\omega \, f(\omega)\right| \leq \mathcal{l}(C_2) M$$
where $M := \mathrm{max}_{\omega \in C_2} |f(\omega)|$ and $\mathcal{l}(C_2) = \pi R$, the contour length. Solving,
$$\frac{d}{d\theta} |f(Re^{-i\theta})| = 0$$
one finds that there is a maximum at $\theta = \pi/2$ and that,
$$|f(Re^{-i\pi/2})| = \frac{\sigma_0}{2\pi} \frac{e^{-Rs}}{|R\tau - 1|}$$
and thus the integral is bounded by,
$$ \left| \int_{C_2} d\omega \, f(\omega) \right| \leq \frac{\sigma_0 R}{2} \frac{e^{-Rs}}{|R\tau-1|}$$
which tends to zero as $R \to \infty$ providing that $s>0$, that is, $t>t'$. Thus the original integral we wanted to evaluate is given by,
$$\int_{-\infty}^\infty \frac{d\omega}{2\pi} \frac{\sigma_0}{1-i\tau \omega}e^{-is\omega} = -\frac{\sigma_0}{2\pi}e^{-(t-t')/\tau}\Theta(t-t').$$
