We can multiply $G$ by a homopoty $H$ to get around the obstacle. More precisely, let $H:[0,1]\times [0,1]\to X$ be the homopoty defined by
$H(s,t)=G(0,st)$. We note that $H(1,t)=G(0,t)=G(1,t)=H(1,t) $ for all $t\in[0,1]$, thus for every $t\in[0,1]$, the product $H_t*G_t*\overline{H_t}$ can be defined, where $H_t:[0,1]\to X, s\mapsto H(s,t)$ is the path of $H$ at time $t$. $\overline{H_t}$ is the reverse path of $H_t$ (i.e., $\overline{H_t}(s)=H_t(1-s)$), the product of paths induces a product of homotopies. We can thus define a homotopy $K:[0,1]\times [0,1]\to X$ by
$$K(s,t)=\begin{cases} H_t(4s)= H(4s,t)=G(0,4st),& s\in[0,1/4]\\
G_t(4s-1)=G(4s-1,t),&s\in[1/4,1/2]\\
\overline{H_t}(2s-1,t)=H(2-2s,t)=G(0,(2-2s)t),&s\in[1/2,1]
\end{cases}$$
Note that $K$ is continuous by the pasting lemma.
Now the path $K(s,0)=H_{0}*G_{0}*\overline{H_0}=e_{x_0}*f*\overline{e_{x_0}}$ is path homotopic to $f$ (note that $[e_{x_0}*f*\overline{e_{x_0}}]=[e_{x_0}]*[f]*[\overline{e_{x_0}}]=[f]$). Similarily, the path $K(s,1)=H_{1}*G_{1}*\overline{H_1}=H_t*e_{x_0}*\overline{H_t}$ is path homotopic to $e_{x_0}$. We also have $K(0,t)=K(1,t)=G(0,0)=x_0$ for all $t\in[0,1]$, thus $K$ is a path homotopy between $e_{x_0}*f*\overline{e_{x_0}}$ and $H_{1}*G_{1}*\overline{H_1}$, thus $e_{x_0}*f*\overline{e_{x_0}}$ and $H_{1}*G_{1}*\overline{H_1}$ are path homotopic, by previous discussion, we see that $f$ and $e_{x_0}$ are also path homotopic.