The normal trick is to set it up as an eigenvalue problem, namely to look instead at $$\Delta u = 0, ~~\frac{\partial u}{\partial n} = \lambda b(x) u(x)~~ \forall x \in \partial D. (1)$$ You first establish that this corresponds to an eigenvalue problem on $L^2(\partial \Omega)$ for a compact operator, thanks to the compact embedding of $H^{1/2}(\partial D)$ into $L^2((\partial D)$.
Then the usual machinery rolls out. There is a discrete spectrum, and there are eigenvalues. And if it so happens that $\lambda=1$ is one of those, then your problem has non trivial solutions.
The advantage of this point of view is that the problem is well posed, there is a clear functional analytic setup, and you don't need much on $b$, some integrability (if it is bounded as you liked to consider, it is fine).
It is done in the following way. Given $f \in H^{-1/2}(\partial D)$ such that $\int_{\partial\Omega} f d\sigma= 0$ consider the so called Neumann-to-Dirichlet map \begin{eqnarray*} NtD : H^{-1/2}(\partial D)/\mathbb R &\to& H^{-1/2}(\partial D)/\mathbb R \\ f &\to& u|_{\partial D} : \begin{cases} -\Delta u =0& \text{ in }D \\ \partial_n u =f &\text{ on }\partial D \end{cases} \end{eqnarray*} the quotient over $\mathbb R$ means that we impose $\int_{\partial\Omega} f d\sigma= \int_{\partial\Omega} u d\sigma= 0.$
The operator $NtD$ is compact, but we are going to use only that it is well defined and continuous.
Now restrict this operator to $L^2(\partial D)/ \mathbb R$, and you can even write it explicitely in terms of Fourier coefficients. $$ f=\sum_{n=1}^\infty a_n \cos n \theta + b_n \sin n \theta, \sum |a_n|^2 + |b_n|^2 <\infty $$ then $$ NtD(f)= \sum_{n=1}^\infty \frac{a_n}{n} \cos n \theta + \frac{b_n}{n} \sin n \theta. $$ It is easy to see that it is a compact operator (the $n^{-1}$ helps).
Supppose $\| b \|_{L^\infty(\partial D)} <\infty$. Then \begin{eqnarray*} B : L^{2}(\partial D)/\mathbb R &\to& L^{2}(\partial D)/\mathbb R \\ f &\to& b f - \frac{1}{\left|\partial D\right|} \int_{\partial D}bf \end{eqnarray*} is a continuous, bounded operator from $ L^{2}(\partial D)/\mathbb R$ into itself.
So the map \begin{eqnarray*} T : L^{2}(\partial D)/\mathbb R &\to& L^{2}(\partial D)/\mathbb R \\ f &\to& B\left(NtD \left(f\right)\right) \end{eqnarray*} is a compact, linear operator. Its spectrum is discrete, it has a maximal eigenvalue etc. and (1) is $$ f=\lambda T f. $$