I am wondering if the solution $u(r) = \ln(r)/\ln(S)$ to the ODE
\begin{align} \Delta_r (u) = \frac{d^2 u}{dr^2} + \frac{1}{r}\frac{\partial u}{\partial r} = 0, \ \ \ u(S) = 1, \ \ \ u(1) = 0, \end{align}
can be expanded in some eigenbasis {$\phi_i$} for the radial Laplacian $\Delta_r$. I've tried considering the expansion $u(r) = \sum_{n=1}^\infty a_n\hat{\phi}_n(r)$, however substituting this into the ODE gives
\begin{align} -\sum_{n=1}^\infty a_n\lambda_n \hat{\phi}_n(r)= 0, \end{align}
which seems to imply that each $a_i = 0$, and therefore no such representation is possible.
Note that the eigenbasis corresponding to the above problem consists of linear combinations of Bessel functions of the first and second kind.
The Sturm-Liouville form of your operator is \begin{align} Lf &= -\frac{d^2f}{dr^2}-\frac{1}{r}\frac{df}{dr}=-\frac{1}{r}\frac{d}{dr}\left(r\frac{df}{dr}\right). \end{align} This is a formally self-adjoint operator on $L^2_r[S,1]$ with respect to the following inner product $$ \langle f,g\rangle_r = \int_S^1f(r)g(r)rdr. $$ However, it is not actually self-adjoint without imposing endpoint conditions such as $f(S)=f(1)=0$. For such conditions imposed on $f,g$, it follows that \begin{align} \langle Lf,g\rangle_r-\langle f,Lg\rangle_r &=-\int_S^1\frac{d}{dr}\left(r\frac{df}{dr}\right)g-f\frac{d}{dr}\left(r\frac{dg}{dr}\right)dr \\ & =-\int_S^1\frac{d}{dr}\left[r\frac{df}{dr}g-rf\frac{dg}{dr}\right]dr \\ & = -r(f'g-fg')|_{S}^{1}=0. \end{align} Therefore $\langle Lf,g\rangle_r = \langle f,Lg\rangle_r$ for $f,g\in\mathcal{D}(L)$. For $S > 0$, this is a regular Sturm-Liouville problem; so the eigenvalues form a discrete set and the corresponding eigenfunctions, once normalized, form a complete orthonormal basis of $L^2[S,1]$. So you'll be able to use this set of eigenfunctions to expand any function in $L^2[S,1]$ using this set of eigenfunctions.