Claim: If $A$ is star convex, then $A$ is simply connected.
Proof: $A$ is clearly path-connected. Let $a\in A$ be the star point, let $\alpha$ and $\beta$ be two loops in $A$, and define $F:I\times I\to A$ by
$$(x,t)\mapsto\begin{cases}(1-2t)\alpha(x)+2ta&t\leq1/2\\2(1-t)a+(2t-1)\beta(x)&t>1/2\end{cases}.$$
Then $F$ is a path homotopy between $\alpha$ and $\beta$, implying that $\pi_1(A,a)=0$.
$$\tag*{$\blacksquare$}$$
Claim: If $\gamma=\alpha*\beta$, then $\widehat\gamma=\widehat\beta\circ\widehat\alpha$.
Proof:
$$\begin{align}\widehat\gamma([f])&=[\overline{\alpha*\beta}]*[f]*[\alpha*\beta]\\&=[\overline\beta*\overline\alpha]*[f]*[\alpha*\beta]\\&=[\overline\beta]*[\overline\alpha]*[f]*[\alpha]*[\beta]\\&=[\overline\beta]*\widehat\alpha([f])*[\beta]\\&=\widehat\beta(\widehat\alpha([f]))\\&=(\widehat\beta\circ\widehat\alpha)([f]).\tag*{$\blacksquare$}\end{align}$$
Claim: $\pi_1(X,x_0)$ is abelian if and only if $\widehat\alpha=\widehat\beta$ for all paths $\alpha,\beta$ from $x_0$ to $x_1$, where $X$ is path-connected.
Proof: Suppose that $\pi_1(X,x_0)$ is abelian and recall that $\pi_1(X,x_1)$ is isomorphic to it. Then
$$\begin{align}\widehat\alpha([f])&=[\overline{\alpha}]*[f]*[\alpha]\\&=[\overline{\alpha}]*[f]*[\beta]*[\overline\beta*\alpha]\\&=[\overline\beta*\alpha]*[\overline{\alpha}]*[f]*[\beta]\\&=[\overline\beta]*[f]*[\beta]\\&=\widehat\beta([f]).\end{align}$$
Conversely, suppose that $\widehat\alpha=\widehat\beta$ for all paths $\alpha,\beta$ from $x_0$ to $x_1$, let $\alpha$ be a path from $x_0$ to $x_1$, let $f$ and $g$ be loops based at $x_0$, and note that $\gamma:=f*\alpha$ is a path from $x_0$ to $x_1$. Then
$$\begin{align}\widehat\gamma([g])&=[\overline\gamma]*[g]*[\gamma]\\&=[\overline{f*\alpha}]*[g]*[f*\alpha]\\&=[\overline\alpha*\overline f]*[g]*[f*\alpha]\\&=[\overline\alpha]*[\overline f*g*f]*[\alpha]\\&=[\overline\alpha]*[g]*[\alpha]\\&=\widehat\alpha([g])\end{align}$$
implies that $[\overline f*g*f]=[g]$, which in turn implies that $[g]*[f]=[f]*[g]$.
$$\tag*{$\blacksquare$}$$
Claim: If $a\in A\subset X$ and $r$ is a retraction of $X$ onto $A$, then
$$r_*:\pi_1(X,a)\to\pi_1(A,a)$$
is surjective.
Proof:
$$r_*\circ\iota_*=(r\circ\iota)_*=\mathrm{id}_{\pi_1(A,a)},$$
where $\iota:A\hookrightarrow X$, implies that $r$ has a right inverse, which in turn implies that it is surjective.
$$\tag*{$\blacksquare$}$$
Claim: If $a\in A\subset\mathbb R^n$, $y\in Y$, $h:\pi_1(A,a)\to\pi_1(Y,y)$, and $h$ is extendable to a continuous $\widetilde h:\mathbb R^n\to Y$, then $h_*$ is trivial.
Proof:
$$h=\widetilde h\circ\iota\implies h_*=\widetilde h_*\circ\iota_*,$$
where $\iota:A\hookrightarrow\mathbb R^n$. However, the domain of $\widetilde h_*$ is $\pi_1(\mathbb R^n,a)=0$.
$$\tag*{$\blacksquare$}$$
Claim: If $X$ is path-connected, $h:X\to Y$ is continuous, $h(x_0)=y_0$, $h(x_1)=y_1$, $\alpha$ is a path from $x_0$ to $x_1$, and $\beta=h\circ\alpha$, then
$$\widehat\beta\circ(h_{x_0})_*=(h_{x_1})_*\circ\widehat\alpha.$$
This is equivalent to saying that $h_*$ is independent of base point up to isomorphism.
Proof:
$$\begin{align}\widehat\beta\circ(h_{x_0})_*([f])&=[\overline\beta]*(h_{x_0})_*([f])*[\beta]\\&=[h\circ\overline\alpha]*[h\circ f]*[h\circ\alpha]\\&=(h_{x_1})_*([\overline\alpha]*[f]*[\alpha])\\&=(h_{x_1})_*\circ\widehat\alpha.\end{align}$$
$$\tag*{$\blacksquare$}$$
Let $G$ be a topological group with operation $\cdot$ and identity $x_0$, let $\Omega(G,x_0)$ be the set of loops in $G$ based at $x_0$, and let
$$(f\otimes g)(s):=f(s)\cdot g(s)$$
for all $f,g\in\Omega(G,x_0)$.
Note that $\Omega(G,x_0)$ equipped with $\otimes$ is a group with identity $x_0(s)=x_0$.
Claim: $\otimes$ induces a group operation $\otimes$ on $\pi_1(G,x_0)$.
Proof: Let $[f]\otimes[g]:=[f\otimes g]$ and note that it is well-defined since $(s,t)\mapsto F(s,t)\cdot G(s,t)$ is a homotopy between $s\mapsto F(s,0)\otimes G(s,0)$ and $s\mapsto F(s,1)\otimes G(s,1)$.
$$\tag*{$\blacksquare$}$$
Claim: $*$ and $\otimes$ on $\pi_1(G,x_0)$ are the same.
Proof: Note that
$$\begin{align}[f]\otimes[g]&=[f*e_{x_0}]\otimes[e_{x_0}*g]\\&=[(f*e_{x_0})\otimes(e_{x_0}*g)]\\&=[f*g]\\&=[f]*[g]\end{align}$$
because
$$\begin{align}((f*e_{x_0})\otimes(e_{x_0}*g))(s)&=(f*e_{x_0})(s)\cdot(e_{x_0}*g)(s)\\&=\begin{cases}f(s)&s\leq1/2\\g(s)&s>1/2\\\end{cases}\\&=(f*g)(s).\end{align}$$
$$\tag*{$\blacksquare$}$$
Claim: $\pi_1(G,x_0)$ is abelian.
Proof:
$$\begin{align}[f]*[g]&=[f]\otimes[g]\\&=[e_{x_0}*f]\otimes[g*e_{x_0}]\\&=[(e_{x_0}*f)\otimes(g*e_{x_0})]\\&=[g*f]\\&=[g]*[f].\end{align}$$
$$\tag*{$\blacksquare$}$$