If the Laplacian operator in spherical coordinates is:
$$ \nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right)$$
then for $a<0$, $\int_{r=0}^\infty e^{ar}\nabla^2\nabla^2 e^{ar} (r^2 dr)$ is
$$\int_0^\infty e^{ar}\nabla^2\left[\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right)e^{ar}\right] (r^2 dr)$$
$$ =\int_0^\infty \color{red}{e^{ar}\nabla^2}\left(a^2 e^{ar} + \frac{2a}{r}e^{ar} \right) (r^2 dr)$$
$$ =? \int_0^\infty\left(a^2 e^{ar} + \frac{2a}{r}e^{ar} \right) \color{red}{\nabla^2 e^{ar}} (r^2 dr)$$
$$ =\int_0^\infty\left(a^2 e^{ar} + \frac{2a}{r}e^{ar} \right)\left(a^2 e^{ar} + \frac{2a}{r}e^{ar} \right) (r^2 dr)$$
$$ =\int_0^\infty\left(a^4 e^{2ar} + \frac{4a^3}{r}e^{2ar} + \frac{4a^2}{r^2}e^{2ar} \right) (r^2 dr)$$
$$ =\int_0^\infty\left(a^4 r^2 e^{2ar} + 4a^3re^{2ar} + 4a^2e^{2ar} \right) dr$$
$$ =\int_0^\infty\left(\frac{a^4}{(2a)^3} \bar r^2 e^{\bar r} + \frac{4a^3}{(2a)^2}\bar r e^{\bar r} + \frac{4a^2}{(2a)}e^{\bar r} \right) d\bar r$$
$$= -a + 2a - 2a$$
$$= -a $$
What is the concrete Hermitian/self-adjoint justification for the inversion of the operator in this equation?