Proof of an identity involving $\int \exp(-|x-s|)dx$ over an even sphere

Following on from the answer of Sam Dolan, I generalized my conjecture to the following, which I prove as Theorem 4 in my paper The magnitude of odd balls via Hankel determinants of reverse Bessel polynomials. [The statement in the question is case where $j=0$.]

Theorem. For $0\le j \le p$, $\mathbf{s}\in \mathbb{R}^{2p+1}$, with $s=|\mathbf{s}|$ and $R>s\ge 0$ \begin{equation*} \int_{\mathbf{x}\in S^{2p}_R} \psi_j(\left | \mathbf{x}-\mathbf{s}\right|)\,\mathrm{d}\mathbf{x} = (-2\pi)^p 2 e^{-R}\sum_{i=0}^{p-j} \binom{p-j}{i}\chi_{i+p}(R) \tilde\psi_{i+j}(s). %\label{eq:GeneralizedKeyIntegral} \end{equation*}

Here $\psi_i$ (as opposed to $\tilde\psi_i$) can be defined by $\psi_i(r)=r^{-2i}e^{-r}\chi_i(r)$; it is related to a modified spherical Bessel function of the second kind.

The sequence of such functions begins in the following way: \begin{align*} \psi_0 (r) &= e^{-r}\\[0.5em] \psi_1 (r) &= e^{-r} \left( \frac{1}{r} \right)\\ \psi_2 (r) &= e^{-r} \left( \frac{1}{r^{2}} + \frac{1}{r^{3}} \right)\\ \psi_3 (r) &= e^{-r} \left( \frac{1}{r^{3}} + \frac{3}{r^{4}} + \frac{3}{r^{5}} \right)\\ \psi_4 (r) &= e^{-r} \left( \frac{1}{r^{4}} + \frac{6}{r^{5}} + \frac{15}{r^{6}} + \frac{15}{r^{7}} \right) . \end{align*}

This theorem is stated in terms of Bessel functions in another question.

Let $L_p$ and $R_p$ be the left and right hand sides. Put $L(z)=\sum_{p=0}^\infty L_p .(z/R)^{2p}$ and similarly for $R(z)$; it will suffice to show that $L(z)=R(z)$. Using the first integral form for $L_p$ one can show that $$ L(z)= \int_{\theta=0}^\pi\frac{z}{2} I_1(\sin(\theta)z) e^{-\sqrt{R^2+s^2-2Rs\cos(\theta)}} \,d\theta, $$ where $I_1$ is a Bessel function. That's not a huge step forwards, but perhaps it is worth something.