Extending to a holomorphic function Let $Z\subseteq \mathbb{C}\setminus \overline{\mathbb{D}}$ be countable and discrete (here $\mathbb{D}$ stands for the unit disc).
Consider a function $f\colon \mathbb{D}\cup Z\to \mathbb{C}$ such that
1) $f\upharpoonright \overline{\mathbb{D}}$ is continuous
2) $f\upharpoonright \mathbb{D}$ is holomorphic
3) if $|z_0|=1$ and $z_n\to z_0$, $z_n\in Z$ then $(f(z_n)-f(z_0))/(z_n-z_0)\to f^\prime(z_0)$
Can $f$ be extended to a holomorphic function on some domain containing $Z$?

No.

The function $g(z) = 1+ 2z + \sum_{n=1}^{\infty} 2^{-n^2} z^{2^n}$ is holomorphic on the open disk $\mathbb{D}$ and infinitely often real differentiable in any point of the closed disk $\overline{\mathbb{D}}$ but cannot be analytically extended beyond $\overline{\mathbb{D}}$:
The radius of convergence is $1$. For $n \gt k$ we have $2^{nk} 2^{-n^2} \leq 2^{-n}$, hence the series and all of its derivatives converge uniformly on $\overline{\mathbb{D}}$, thus $g$ is indeed smooth in the real sense and holomorphic on $\mathbb{D}$. By Hadamard's theorem on lacunary series the function $g$ cannot be analytically continued beyond $\overline{\mathbb D}$.
[In fact, it is not difficult to show that $g$ is injective on $\mathbb{\overline{D}}$, so $g$ is even a diffeomorphism onto its image $g(\overline{\mathbb D})$ — but that's not needed here.]
I learned about this nice example from Remmert, Classical topics in complex function theory, Springer GTM 172, Chapter 11, §2.3 (note: I'm quoting from the German edition). The entire chapter is devoted to the behavior of power series on the boundary of convergence and gives theorems that provide positive and negative answers on whether a given power series can be extended or not.

Now, to get a counterexample, apply Whitney's theorem to extend $g$ to a smooth function $f$ on all of $\mathbb{C}$. Every restriction of $f$ to any set of the form $\overline{\mathbb{D}} \cup Z$ (I think that's what was intended) as in your question will provide a counterexample.
