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 $k$.
I notice that your derivatives $f^{(n)}(x)$ are off by an index; more importantly, there is a simplification. For $n\geq1$, we have \begin{align*} f^{(n)}(x)&=\frac{2}{\sqrt{\pi}}e^{(a+x)^2}\sum_{m=0}^n\sum_{j=0}^{m}\frac{\binom{m}{j}(-1)^j(a+x)^{2m-n}}{m!} \prod_{p=1}^n(2m-2j-p+1) \\ &=\frac2{\sqrt{\pi}}e^{(a+x)^2}(a+x)^{2m-n}\sum_{m=0}^n\binom{m}{n-2m}\frac{m!n!}{2^{n-2m}}. \end{align*}