I've been trying to condense the sequence of steps for arguing that there is no elementary form for the integral as follows:
Consider DE's over $\mathbb{C}(x)$ with the usual derivation.
Let the solutions to $L=y''+x^ny'=0 $ be $1$ and $z=\int{e^{-\int{x^n}}}$ (i.e. $z'=e^{-\int{x^n}}$ and $z''=-x^n\,e^{-\int{x^n}}$, satisfying the DE).
We aim to show this only has an elementary solution when $n=-1$ (i.e. that $G(\mathbb{C}(x)\langle1,z\rangle)/\mathbb{C}(x))$ is only soluble in this case).
A basis for the permutations of solutions could consist of mapping $1$ to $1$ along with $z$ to $c+dz$.
If $\sigma$ is a differential automorphism, $\sigma(z')=\sigma(z)'=(c+dz)'=dz'$.
The automorphisms can be represented by the set of matrices $\begin{pmatrix}1 & c \\ 0 & d\end{pmatrix}$ in $GL(2,C)$.
Two such matrices only commute in general if $d=1$ (that they should commute follows from the connected component of the identity being abelian if the Galois group is soluble).
So we have that $\sigma(z')=z'$ with $z'=e^{-\int{x^n}}$ as at the start.
This means that $z'=e^{-\int{x^n}}$ must be in $\mathbb{C}(x)$ as it is mapped to itself. [My main question - is this a valid inference?].
Hence $e^{-\int{x^n}} = c$ for some polynomial in $\mathbb{C}(x)$, i.e. $-\int{x^n}=ln(c)$ which only happens when $n=-1$ and $c$ only has an $x$ term.
If this is all valid then there is no elementary solution to $y''+xy'=0$ as $n=1$ in this equation, and hence there is no elementary form for $\int e^{\frac{x^2}{2}}\,dx$, i.e. none for $\int e^{x^2}\,dx$.
It turns out that the above is only a good summary of why the anti-derivative of $e^{x^2}$ is not elementary if we are using A. Magid's definition of elementary in his 'Lectures on Differential Galois Theory'. As mentioned here: Elementary function defined by A Magid Magid does not regard this definition as capturing the usual, intuitive definition of elementary. So, while perhaps of interest, the derivation I originally posted here is not really the best explanation of why the anti-derivative of $e^{x^2}$ is not elementary, the better explanation being Liouville's original work.