(This question is originally from Math.SE where it was suggested that I ask the question here)

Let $S$ be a simply connected subset of $\mathbb{R}^2$ and let $x$ be an interior point of $S$, meaning that $B_r(x)\subseteq S$ for some $r>0$. Is it necessarily the case that $\pi_1(S\setminus\{x\})\cong\mathbb{Z}$?

This is true if $S$ is open by the Riemann mapping theorem.
Moishe Kohan's answer on Math.SE shows that the result holds if $S$ is compact.

Let $B=B_r(x)$ and let $G=\pi_1(S\setminus\{x\})$. I can show that $G^{ab}\cong\mathbb{Z}$. Consider the commutative diagram of topological spaces $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c}B\setminus\{x\}&\ra{}&S\setminus\{x\}&\ra{}&\mathbb{R}^2\setminus\{x\}\\\da{}&&\da{}\\B&\ra{}&S\end{array} $$ Applying the fundamental group functor gives a commutative diagram of groups $$ \begin{array}{c}\mathbb{Z}&\ra{}&G&\ra{}&\mathbb{Z}\\\da{}&&\da{}\\1&\ra{}&1\end{array} $$ where the composition of the top two maps is the identity homomorphism on $\mathbb{Z}$. By the Seifert-van Kampen theorem, the square is a pushout diagram of groups, meaning that the normal closure of the image of $\mathbb{Z}\to G$ is all of $G$. In particular, if $A$ is an abelian group then any two homomorphisms $G\to A$ that agree on the image of $\mathbb{Z}\to G$ must agree on all of $G$. Another way to put this is that if we have two homomorphisms $G\to A$ such that the two compositions $\mathbb{Z}\to G\to A$ are equal then the two homomorphisms $G\to A$ are equal.

I claim that the map $G\to\mathbb{Z}$ is an abelianization map. To see this, let $A$ be an abelian group and let $G\to A$ be a homomorphism. Now recall that the composition $\mathbb{Z}\to G\to\mathbb{Z}$ is the identity. Then the composition $\mathbb{Z}\to G\to\mathbb{Z}\to G\to A$ agrees with the composition $\mathbb{Z}\to G\to A$. By the remark at the end of the previous paragraph, this means that the composition $G\to\mathbb{Z}\to G\to A$ agrees with the map $G\to A$. In other words, the composition $\mathbb{Z}\to G\to A$ makes the abelianization diagram commute.

To show uniqueness, let $\mathbb{Z}\to A$ be a map making the abelianization diagram commute. Then the composition $G\to\mathbb{Z}\to A$ agrees with the map $G\to A$. Then the composition $\mathbb{Z}\to G\to\mathbb{Z}\to A$ agrees with the composition $\mathbb{Z}\to G\to A$. Since the composition $\mathbb{Z}\to G\to\mathbb{Z}$ is the identity, this shows that the map $\mathbb{Z}\to A$ is given by the composition $\mathbb{Z}\to G\to A$.

This shows that the map $G\to\mathbb{Z}$ is an abelianization map.