In this answer stability is covered first, thereafter the effect of the PID-controller on the stability.
Consider the following simple control system
which has a closed loop transfer function of $$\Gamma(s)= \frac{C(s)G(s)}{1+C(s)G(s)}=\frac{L(s)}{1+L(s)}.$$
Here $L(s) = C(s)G(s)$ is called the loop.
The system is unstable if at frequency $\omega$ where $\angle L(j\omega) = \arg\left(L(j\omega)\right)=-180^\circ$, the magnitude $|L(j\omega)|>1(=0$dB$)$.
Similarly, the system is stable if at frequency $\omega$ where $\angle L(j\omega) = \arg\left(L(j\omega)\right)=-180^\circ$, the magnitude $|L(j\omega)|<1(=0$dB$)$.
Furthermore, neutral (or marginal) stability is at frequency $\omega$ where $\angle L(j\omega) = \arg\left(L(j\omega)\right)=-180^\circ$, the magnitude $|L(j\omega)|=1(=0$dB$)$.
The stability margin is ''the space'' you have until the system becomes unstable (i.e. robustness).
Since stability is determined by using two measures: the gain and the phase, consequently there are also two stability margins (the gain margin and the phase margin).
The gain margin is ``the space'' you have at frequency $\omega$ where $\angle L(j\omega) = \arg\left(L(j\omega)\right)=-180^\circ$ until $|L(j\omega)| = 0$.
The phase margin is ``the space'' you have at frequency $\omega$ where $|L(j\omega)| = 0$ until $\angle L(j\omega) = \arg\left(L(j\omega)\right)=-180^\circ$ until.
The gain, phase and corresponding margins can be computed by hand, however I would recommend using software (e.g. MATLAB), since they can be very extensive.
To find effect your controller $C(s)$, you can draw Bodediagram and see the effect of the various parameters (MATLAB command bode()):
The margins of the complete system can also be visualized using MATLAB (command margin()), which should give more insight in the effect of the chosen controller and your stability margins.
