While the quoted text above is correct and gives a worded explanation on how angular momentum is conserved in a two-body central force problem, generally, more rigor is required for a proof. We know that given the Lagrangian of the system the generalized momentum is defined as:
$$
p_j = \frac{\partial{L}}{\partial{\dot{q_j}}}
$$
and also the Lagrange equations of motion are expressed by:
$$
\dot{p}_j=\frac{\partial{L}}{\partial{q_j}}
$$
The Lagrangian for a two-body system (reduced to a one-body problem) of reduced mass $\mu$ moving in a central-force field described by the potential function $U(r)$ in polar coordinates is:
$$
L=\frac{1}{2}\mu(\dot{r}^2+r^2\dot{\theta}^2)-U(r)
$$
from here we can take our angular momentum conjugate to $\theta$ as
$$
\dot{p}_\theta = \frac{\partial L}{\partial \theta}=0=\frac{d}{dt}\frac{\partial{L}}{\partial{\dot{\theta}}}
$$
since the Lagrange equation of motion in $\theta$ is
$$
\frac{\partial L}{\partial \theta}-\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta}}=0
$$
therefore
$$
\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta}}=\frac{d}{dt}(p_\theta)=\texttt{constant}
$$
Hence, angular momentum is conserved. And this all relies on the fact that $\theta$ is a cyclic coordinate which made its conjugate momentum zero as the quoted text states.