On every simply connected domain, there exists a holomorphic function with no analytic continuation.

I don't think this has much to do with simple connectivity. Let $U\subset \mathbb {C}$ be any open set with non-empty boundary. Claim: There exists $f\in H(U)$ such that

$$\sup_{D(a,r)\cap U} |f|=\infty$$

for every $a \in \partial U$ and $r>0.$ This $f$ cannot be be extended analytically to any larger open set containing a point in $\partial U.$

Proof: Let $\{a_1,a_2, \dots \}$ be a countable dense subset of $\partial U.$ The idea is to define

$$f(z) = \sum_{n=1}^{\infty} \frac{c_n}{z-a_n}$$

for appropriate positive constants $c_n.$

To do this, we choose pairwise disjoint countably infinite sets $E_1, E_2,\dots \subset U$ such that for each $n,$ the only limit point of $E_n$ in $\mathbb {C}$ is $a_n.$ (For each $n,$ $E_n$ is just a sequence of distinct points in $U$ whose limit is $a_n.$ You can choose the $E_n$'s inductively.)

Now there are compact sets $K_1,K_2, \dots \subset U$ such that

$$K_1\subset \text {int}(K_2)\subset K_2 \subset \text {int}(K_3) \subset \dots$$

such that $U= \cup K_n.$ Choose $c_1$ such that $c_1/|z-a_1| < 1/2$ on $K_1.$ If $c_1,\dots c_n$ have been chosen, we choose $c_{n+1}$ such that

$$\frac{c_{n+1}}{|z-a_{n+1}|} < \frac{1}{2^{n+1}},\ \ z\in K_{n+1}\cup E_1 \cup \cdots E_n.$$

Then the series $\sum c_n/(z-a_n)$ converges uniformly on each $K_n,$ hence defines a holomorphic function $f$ on $U.$ Let $a\in \partial U, r > 0.$ Then $D(a,r)$ contains some $a_n,$ hence contains the tail-end of $E_n.$ Thus as $z\to a_n$ within $E_n,$ we have

$$|f(z)| \ge \left|\sum_{k=1}^{n}\frac{c_n}{z-a_k}\right| - \sum_{k=n+1}^{\infty}\frac{1}{2^{k+1}}\to \infty.$$

This completes the proof.