To avoid "extended discussion", here is what I said.
\begin{align} (f(x).g(x))^{(n)} &=\sum_{k=0}^{n}\binom{n}{k}f^{(k)}(x)g^{(n-k)}(x)\\
&=f(x)g^{(n)}+\sum_{k=1}^{n}\binom{n}{k}f^{(k)}(x)g^{(n-k)}(x) \\
&=\frac2{\sqrt{\pi}}c^ne^{cx}\int_0^{a+x}e^{t^2}dt
+\sum_{k=1}^{n}\binom{n}{k}f^{(k)}(x)g^{(n-k)}(x).
\end{align}
Now use your formulas for $f^{(k)}(x)$, for $k\geq1$, and the obvious one $g^{(k)}(x)$ for any $k$.

there is a simplification of your bouble-sum. For $n\geq1$, we have
\begin{align*}
f^{(n)}(x)&=\frac{2}{\sqrt{\pi}}e^{(a+x)^2}\sum_{m=0}^{n-1}\sum_{j=0}^{m}\frac{\binom{m}{j}(-1)^j(a+x)^{2m-n+1}}{m!}
\prod_{p=1}^{n-1}(2m-2j-p+1) \\
&=\frac{2}{\sqrt{\pi}}e^{(a+x)^2}(a+x)^{2m-n+1}\sum_{m=0}^{n-1}\frac{(n-1)!}{m!}\sum_{j=0}^{m}(-1)^j\binom{m}j\binom{2m-2j}{n-1}\\
&=\frac2{\sqrt{\pi}}e^{(a+x)^2}(a+x)^{2m-n+1}(n-1)!\sum_{m=0}^{n-1}\frac1{m!}\binom{m}{n-m-1}\frac1{2^{n-2m-1}}.
\end{align*}
So, here is the identity I used:
$$\sum_{j=0}^{m}(-1)^j\binom{m}j\binom{2m-2j}n=\binom{m}{n-m}\frac1{2^{n-2m}}.$$
As far as proving this and similar identities goes, you may follow the procedure I outlined [in my answer here][1].

[1]: https://mathoverflow.net/questions/193611/binomial-coefficient-identity/257970#257970