This is known as the $\epsilon$ prescription. We have:
\begin{equation}
\begin{aligned}
\tilde{G}(E) =& - \frac{i}{\hbar} \sum_{a^{'}} \int^{\infty}_0 dt \exp\left [ \frac{it (E - E_{a'})}{\hbar} \right]\\
=& - \frac{i}{\hbar} \sum_{a^{'}} \frac{\hbar}{i} \left( \frac{1}{E - E_{a'}} \right) \exp\left [ \frac{it (E - E_{a'})}{\hbar} \right] \Bigg|_0^{\infty}
\end{aligned}
\end{equation}
Now, the problem is that we cannot evaluate the upper limit of the integral because
$$\lim_{t \to \infty}\exp (itx)$$
does not exist for real $x$. The limit does not converge to a definite value; it keeps oscillating instead.
The trick to force convergence upon the integral is to add to $E$ an infinitesimal constant $\epsilon > 0$:
$$ E \rightarrow E + i \epsilon$$
Now, the upper limit of the integral goes to zero because of exponential suppression:
\begin{equation}
\lim_{t \to \infty}- \frac{i}{\hbar} \sum_{a^{'}} \frac{\hbar}{i} \left( \frac{1}{(E + i \epsilon) - E_{a'}} \right) \exp\left [ \frac{it (E - E_{a'})}{\hbar} \right] \exp \left[ \frac{-\epsilon t}{\hbar} \right] = 0
\end{equation}
while the lower limit (with minus sign included) gives:
\begin{equation}
\lim_{t \to 0} \frac{i}{\hbar} \sum_{a^{'}} \frac{\hbar}{i} \left( \frac{1}{(E + i \epsilon) - E_{a'}} \right) \exp\left [ \frac{it (E - E_{a'})}{\hbar} \right] \exp \left[ \frac{-\epsilon t}{\hbar} \right] = \sum_{a^{'}} \left( \frac{1}{(E + i \epsilon) - E_{a'}} \right)
\end{equation}
The $\epsilon$ prescription was used for convergence of the integral. The integral has now been done so we can safely take $\epsilon \to 0$ in the end to get:
$$\tilde{G}(E) = \sum_{a^{'}} \frac{1}{E - E_{a'}}$$
Analytic continuations such as this $\epsilon$ prescription are used in many places in physics. One of the key motivations is convergence.
