Hypothesis: Let $p$ be a non-constant $\mathbb{C}$-polynomial, and let $$s_{p}(z):=\sum_{\zeta : p(\zeta)=z}{\frac{f(\zeta)}{p'(\zeta)}}\qquad(z\in U)$$ be its formal Stieltjes transform, where $U\subset\mathbb{C}$ is a non-empty open subset such that for any $z\in U$, $f$ is analytic around $\zeta\in\mathbb{C}$ satisfying $p(\zeta)=z$.
To prove: If the set $\Omega:=\{z\in U:\exists\zeta\in\mathbb{C},p(\zeta)-z=0=p'(\zeta)\}$ is finite, then the elements of $\Omega$ are removable singularities of $s_{p}$.
My attempt: Let $z_{o}\in\Omega$ be arbitrary and let $\zeta_{o}\in\mathbb{C}$ be such that $p(\zeta_{o})-z_{o}=0=p'(\zeta_{o})$. Since $\Omega$ is finite, then we can find a path $\gamma\subset\mathbb{C}$ which encloses $\zeta_{o}$, and so that $f/p'$ is analytic on it and on its interior (except at $\zeta_{o}$, of course). Therefore, for $\zeta$ close enough to $\zeta_{o}$, $$(\zeta-\zeta_{o})\frac{f(\zeta)}{p'(\zeta)}=\frac{(\zeta-\zeta_{o})^{2-n_{o}}}{2\pi\text{i}q(\zeta)}\oint_{\gamma}{\frac{f(x)}{x-\zeta}\mathrm{d}x}$$ by the analytic nature of $f$, where $n_{o}$ is the multiplicity of $\zeta_{o}$ (w.r.t $p$) and $q$ is a non-vanishing polynomial on a small enough neighborhood of $\zeta_{o}$ enclosed by $\gamma$.
My idea is to use Riemann's theorem to conclude, but it fails for $n_{o}\ge 2$.
Edit: The author just suggested me: "Try using contour integrals of $\frac{f(\zeta)}{p'(\zeta)}$ instead." But once again, I failed to conclude. I obtained, for instance, that the coefficients $b_{n}$ ($n\ge 1$) of the principal part of the Laurent expansion of $\frac{f(\zeta)}{p'(\zeta)}$ are given by $$\oint_{\gamma}{\frac{f(x)}{q(x)}(x-\zeta_{o})^{n-n_{o}}\frac{\mathrm{d}x}{2\pi\text{i}}}.$$ Thus, by the analytic nature of $\frac{f(x)}{q(x)}$ at $\zeta_{o}$, for all $n\ge n_{o}$, $b_{n}=0$. By the way, this is the final part of the proof of Prop. 2.5.7 on Topics in Random Matrix Theory by Terence Tao (https://terrytao.files.wordpress.com/2011/08/matrix-book.pdf).
New attempt: I would like to know if the following is correct: $$\frac{f(\zeta_{o})}{p'(\zeta_{o})}:=\oint_{\gamma}{\frac{f(x)/p'(x)}{x-\zeta_{o}}\frac{\mathrm{d}x}{2\pi\text{i}}}=\oint_{\gamma}{\frac{f(x)/q(x)}{(x-\zeta_{o})^{n_{o}}}\frac{\mathrm{d}x}{2\pi\text{i}}}=\frac{1}{(n_{o}-1)!}\frac{\mathrm{d}^{(n_{o}-1)}}{\mathrm{d}x^{(n_{o}-1)}}\Bigg[\frac{f(x)}{q(x)}\Bigg]_{x=\zeta_{o}}$$ is well-defined since $q$ does not vanish on a small enough neighborhood of $\zeta_{o}$. Is this enough to claim that $\zeta_{o}$ is a removable singularity?. Sorry in advance if this is a silly question.