The graph of $\mathcal{G}(A)$ is closed. The orthogonal complement of $\mathcal{G}(A)$ consists of all pairs $(y,z)\in\mathcal{H}\times\mathcal{H}$ such that
$$
\langle (x,Ax),(y,z)\rangle =0,\;\; x\in\mathcal{D}(A).
$$
That is the same as
$$
\langle x,y\rangle+\langle Ax,z\rangle=0,\;\; x\in\mathcal{D}(A),
$$
which is exactly the requirement that $z\in\mathcal{D}(A^*)$ and $A^*z=-y$. Therefore, every $(z,w)\in\mathcal{H}\times\mathcal{H}$ may be written as the following sum for some $x\in\mathcal{D}(A)$ and $y\in\mathcal{D}(A^*)$:
$$
(z,w) = (x,Ax)+(-A^*y,y)
$$
Therefore,
$$
z=x-A^*y,\;\; w=Ax+y
$$
We can choose $w=0$ in order to obtain $x,y$ such that
$$
z=x-A^*y=x+A^*Ax.
$$
It follows that $A^*A+I$ is surjective because for every $z$ there exists $x$ such that $z=(I+A^*A)x$. It is easy to verify that $A^*A+I$ is symmetric on its natural domain.
The domain of $A^*A+I$ has to be dense because it is surjective; to see why, suppose $y\perp\mathcal{D}(A^*A)$. Then there exists $x\in\mathcal{D}(I+A^*A)$ such that $(I+A^*A)x=y$. Because $y\perp x$, it follows that
$$
0= \langle y,x\rangle = \langle (I+A^*A)x,x\rangle=\|x\|^2+\|Ax\|^2\implies x=0 \implies y = 0.
$$
So $I+A^*A$ is densely-defined, symmetric and surjective.
That is enough to conclude that $I+A^*A$ is self-adjoint. Indeed, suppose $y\in\mathcal{D}((I+A^*A)^*)$. Then
$$
\langle (I+A^*A)x,y\rangle = \langle x,(I+A^*A)^*y\rangle,\;\; x\in\mathcal{D}(I+A^*A).
$$
Then there exists $w\in\mathcal{D}(I+A^*A)$ such that $(I+A^*A)^*y=(I+A^*A)w$. Hence,
$$
\langle (I+A^*A)x,y\rangle = \langle x,(I+A^*A)^*y\rangle= \langle x,(I+A^*A)w\rangle = \langle (I+A^*A)x,w\rangle
$$
Because $I+A^*A$ is surjective, then $y=w$, from which it follows that $y\in\mathcal{D}(I+A^*A)$ and $(I+A^*A)^*y=(I+A^*A)y$. Therefore, $(I+A^*A)^*=I+A^*A$.