By dlmf 8.4.6
$$
\Gamma\left(\tfrac{1}{2},z\right)=2\int\limits_{\sqrt{z}}^{\infty}e^{-t^{2}}dt.
$$
For negative arguments we have
$$
\Gamma\left(\tfrac{1}{2},-|z|\right)=2\int\limits_{i\sqrt{|z|}}^{\infty}e^{-t^{2}}dt=2\int\limits_{i\sqrt{|z|}}^{i\sqrt{|z|}\,+\infty}e^{-t^{2}}dt=2\int_{0}^{+\infty}e^{-\left(t+i\sqrt{|z|}\right)^{2}}dt.
$$
$$
\text{Re}\left\{\Gamma\left(\tfrac{1}{2},-|z|\right)\right\}=2\int_{0}^{+\infty}e^{-\left(t^2-|z|\right)}\cos\left(2t\sqrt{|z|}\right)dt=2e^{|z|}\cdot\frac{\sqrt{\pi}}{2}e^{-|z|}=\Gamma\left(\frac12\right).
$$
By dlmf 8.8.2
$$
\Gamma\left(a+1,z\right)=a\Gamma\left(a,z\right)+z^{a}e^{-z}.
$$
From this one can easily deduce by induction that the claim is true for all $n\in\mathbb{Z}$:
$$\text{Re}\left\{\Gamma\left(n+\tfrac{1}{2},-|z|\right)\right\}=\Gamma\left(n+\tfrac{1}{2}\right).$$
Numerical checked confirmed this formula for non-negative $n$ too.
$Q(t) = \frac{e^{at}}{a}\Big[e^{b/a}-e^{-at}\sum_{k=0}^\infty \sum_{l=0}^k \frac{(at)^l(b/a)^k}{k!l!}\Big].
$
I'll blindly try
to reverse the order of summation
and see what happens.
$\begin{array}\\
S(u, v)
&=\sum_{k=0}^\infty \sum_{l=0}^k \frac{u^lv^k}{k!l!}\\
&=\sum_{l=0}^\infty\sum_{k=l}^\infty \frac{u^lv^k}{k!l!}\\
&=\sum_{l=0}^\infty\frac{u^l}{l!}\sum_{k=l}^\infty \frac{v^k}{k!}\\
&=\sum_{l=0}^\infty\frac{u^l}{l!}(e^v-\sum_{k=0}^{l-1} \frac{v^k}{k!})\\
&=\sum_{l=0}^\infty\frac{u^l}{l!}e^v-\sum_{l=0}^\infty\frac{u^l}{l!}\sum_{k=0}^{l-1} \frac{v^k}{k!}\\
&=e^ue^v-\sum_{l=0}^\infty\frac{u^l}{l!}\sum_{k=0}^{l-1} \frac{v^k}{k!}\\
&=e^{u+v}-\sum_{l=0}^\infty\frac{u^l}{l!}(\sum_{k=0}^{l} \frac{v^k}{k!}-\frac{v^l}{l!})\\
&=e^{u+v}-\sum_{l=0}^\infty\frac{u^l}{l!}\sum_{k=0}^{l} \frac{v^k}{k!}+\sum_{l=0}^\infty\frac{u^l}{l!}\frac{v^l}{l!}\\
&=e^{u+v}-\sum_{l=0}^\infty\sum_{k=0}^{l}\frac{u^l}{l!} \frac{v^k}{k!}+\sum_{l=0}^\infty\frac{(uv)^l}{l!^2}\\
&=e^{u+v}-S(v, u)+I_0(2\sqrt{uv})
\\
\end{array}
$
where
$I_0$
is the modified Bessel function
of the first kind.
So this isn't a evaluation
but we get the relation
$S(u, v)+S(v, u)
=e^{u+v}+I_0(2\sqrt{uv})
$.
Then
$\begin{array}\\
Q(t)
&= \frac{e^{at}}{a}\Big[e^{b/a}-e^{-at}\sum_{k=0}^\infty \sum_{l=0}^k \frac{(at)^l(b/a)^k}{k!l!}\Big]\\
&= \frac{e^{at}}{a}\Big[e^{b/a}-e^{-at}S(at, b/a)\Big]\\
&= \frac{1}{a}\Big[e^{at+b/a}-S(at, b/a)\Big]\\
&= \frac{1}{a}\Big[e^{at+b/a}-(e^{at+b/a}-S(b/a, at)+I_0(2\sqrt{(at)(b/a)}))\Big]\\
&= \frac{1}{a}\Big[S(b/a, at)-I_0(2\sqrt{tb})\Big]\\
\end{array}
$
Again,
not an evaluation,
but a possibly useful
alternative expression.
This reminds me
very much
of some work I did
over forty years ago
on the Marcum Q-function.
You might look that up
and follow the references.
You can start here:
https://en.wikipedia.org/wiki/Marcum_Q-function
Best Answer
There is a sum formula for negative $n$ but it contains the exponential integral $E_1(x)$ given in http://dlmf.nist.gov/8.4#E15: $$\Gamma(-n,x) = \frac{(-1)^n}{n!}\left( E_1(x) - e^{-x} \sum_{k=0}^{n-1} \frac{(-1)^k k!}{x^{k+1}}\right)$$