Let $E \subset \mathbb{C}$ be a compact set. Its logarithmic capacity is defined as $$ \DeclareMathOperator{\capacity}{cap} \capacity(E) := \exp\left(-\inf_\mu \iint \log\frac{1}{|z-w|} \, d\mu(z) \, d\mu(w)\right), $$ where the infimum is taken over all positive unit Borel measures supported on $E$. A compact set $P \subset \mathbb{C}$ is called polar if $\capacity(P) = 0$.
Does $\capacity(E) = \capacity(E \cup P) = \capacity(E \setminus P)$ hold?
Yes. I'll prove $\DeclareMathOperator{\capacity}{cap} \capacity(E) = \capacity(E \cup P)$, and the other case can be treated analogously. W.l.o.g., we can assume $E \cap P = \{\}$. Let $\mu^\star$ be the minimising measure for $E \cup P$. (This measure always exists, and is unique if $\capacity(E\cup P) > 0$.) We can split $\mu^\star = \mu_E + \mu_P$ into parts supported on $E$ and $P$. If $\mu_P = 0$, it follows $\capacity(E \cup P) \leq \capacity(E)$ and since $\capacity(E \cup P) \geq \capacity(E)$ for any $P$ (the infimum is taken over a larger set on the right-hand side), the claim follows. If $\mu_P \neq 0$, we have \begin{align} \capacity(E \cup P) &= \exp\left(-\iint \log\frac{1}{|z-w|} \, d\mu^\star(z) \, d\mu^\star(w)\right) \\&= \exp\left( -\underbrace{\iint \log\frac{1}{|z-w|} \, d\mu_E(z) \, d\mu_E(w)}_{> -\infty} \right.\\ &\qquad\qquad -\underbrace{2\iint \log\frac{1}{|z-w|} \, d\mu_E(z) \, d\mu_P(w)}_{> -\infty} \\ &\qquad\qquad \left. -\underbrace{\iint \log\frac{1}{|z-w|} \, d\mu_P(z) \, d\mu_P(w)}_{=\infty} \right) \\ &= 0 \end{align} because $E$ and $P$ are bounded and $P$ is polar. Since $\capacity(E \cup P) \geq \capacity(E)$, the claim follows.