Let us just use your second identity, $$ J_0(pr)+\sum_{n=1}^\infty b_n J_{\pi n/ \phi} (pr)=1 \mbox{ for all }r\in(0,r_0) $$ Since $J_{\nu}(x)\approx \left(\frac{z}{2}\right)^{\nu}\frac{1}{\Gamma(\nu+1)}$, and $$ 1-J_0(x)=\sum_{k=1}^\infty \left(\frac{z^{k}}{2 (k!)}\right)^2(-1)^{k+1} $$ you obtain at first order $$ b_1 \left(\frac{pr}{2}\right)^{\pi/\phi}\frac{1}{\Gamma(\pi/\phi+1)} \approx \left(\frac{pr}{2}\right)^{2} $$
so if $\pi/\phi\neq 2$ there are no solutions. If $2\phi=\pi$, $b_1=2$ (and as you pointed out all other $b_i$ as well).

Let us just use your second identity, $$ J_0(pr)+\sum_{n=1}^\infty b_n J_{\pi n/ \phi} (pr)=1 \mbox{ for all }r\in(0,r_0) $$ Since $J_{\nu}(x)\approx \left(\frac{z}{2}\right)^{\nu}\frac{1}{\Gamma(\nu+1)}$, and $$ 1-J_0(x)=\sum_{k=1}^\infty \left(\frac{z^{k}}{2 (k!)}\right)^2(-1)^{k+1} $$ you obtain at first order $$ b_1 \left(\frac{pr}{2}\right)^{\pi/\phi}\frac{1}{\Gamma(\pi/\phi+1)} \approx \left(\frac{pr}{2}\right)^{2} $$
so if $\pi/\phi\neq 2$ there are no solutions. If $2\phi=\pi$, $b_1=2$ (and as you pointed out all other $b_i$ as well). Added after comments: from the growth of $J_{\nu}$ given above, you can see that this is also the answer provided $$ b_n < (C)^{\pi n/\phi}\Gamma(\frac{\pi n}{\phi}+1) $$ for some positive $C$, provided that $2Cr_0\leq 1$. Now if you want an approximate solution, that's a completely different question, but then you have to explain which type (in which norm) you want the approximation to hold...and forget uniqueness.