 Solution

Choose the common initial longitude of the particle and $G$ as zero. Let the fixed inertial longitude of periapsis be $\varpi$, positive ahead of that ray. At release $f_0=-\varpi$, but the mean anomaly isThereafter $M(t)=nt+M_0$, $f(t)=M(t)+2e\sin M(t)+O(e^2)$, and the particle's longitude is $\lambda (t)=\varpi +f(t)$. The reference ray has longitude ${\lambda }_G=n_{g}t$.

In the rotating reference frame, the radial $x$ axis points outward along the star-$G$ ray and the $y$ axis points forward. Define $\delta \lambda =\lambda -{\lambda }_G$. The exact positional relations are

The diagram shows a later instant; initially $\delta \lambda =0$, so $y(0)=0$. Retaining the $O(e)$ correction in $M_0$ is what will preserve that initial condition in the approximate trajectory.