Inverse Fourier of Two-Pole Transfer Function
The comments to the question already hint at the solution, but I'll write it out.

$\tilde{H}(\omega)$ can converted from a product into a sum using Partial Fraction Expansion:

$$\tilde{H}(\omega) = \frac{1}{1 + i \omega \tau_1} \frac{1}{1 + i \omega \tau_2} = \frac{A}{1 + i \omega \tau_1} +\frac{B}{1 + i \omega \tau_2}$$

$$ 1 \omega ^0 + 0 \omega ^1 = A(1 + i \omega \tau_2) + B(1 + i \omega \tau_1)$$

Which gives
$$\implies A + B = 1$$
and
\begin{equation}
\begin{split}
A i \omega \tau_2 + B + i \omega \tau_1 &amp;= 0\\
(1 - B) \tau_2 + B \tau_1 &amp;= 0\\
B(\tau_1 - \tau_2) &amp;= -\tau_2\\
B = \frac{\tau_2}{\tau_2 - \tau_1}
\end{split}
\end{equation}

$$A = 1 - B = \frac{- \tau_1}{\tau_2 - \tau_1}$$

Leading to:

$$\tilde{H}(\omega) = \Big(\frac{1}{\tau_2 - \tau_1}\Big) \Big[ \frac{\tau_2}{1 + i \omega \tau_2} -\frac{\tau_1}{1 + i \omega \tau_1}\Big]$$

$\tilde{H}(\omega)$ also contains a term $e^{-i\omega t_d}$, so now for each of the terms in the sum above the inverse Fourier transform is:

$$\mathcal{F}^{-1}\Big\{\Big(\frac{\tau}{1 + i \omega \tau}\Big) e^{-i\omega t_d}\Big\} = \Theta(t) ~ e^{\displaystyle -\frac{t - t_d}{\tau}} $$

and their sum has the desired form in the question.