Some comments from my side:
1.) The stability check in the BODE diagram concerns the LOOP GAIN response only (because once you did mention "closed-loop system" in your text.)
2.) The shown system is "conditionally stable". That means: It is stable - regardless the properties at the frequency A. However, if you REDUCE the gain within the loop until the gain crosses the point A (the phase remains unchanged) the closed-loop system will be unstable.
Such conditional stable system should be avoided because a gain reduction can happen due to aging or other damping effects. Remember: Classical feedback systems with a continuos decreasing loop phase will become unstable (under closed-loop conditions) for rising gain values (beyond a certain limit) only.
As to your next question - the input signal Vi does not influence stability properties at all. Stability is determined by the loop components only. That is the reason, we investigate the loop gain only.
EDIT: Here is an explanation why the closed loop (your example) will be stable:
If a closed-loop system is unstable, this point of instability also must be "stable". That means - either we will have "stable" and continuous oscillations or the output is latched at one of the supply voltage rails. In both cases, this point of instability is fixed.
Now - what happens at the point A in your example? Here we have a rising phase which is identical to a NEGATIVE group delay at this point (group delay is defined as the negative phase slope). This is an indication for the unability of the closed-loop system to let the amplitudes rise (oscillations or latching at the supply rail). Rather, the system returns to a stable operating point.
A final information: The stability check investigates either (a) the -180deg line or (b) the -360 deg line. This depends on what you are investigating: (a) Either the simple product GH or (b) the loop gain LG which is LG=-GH.
If there exists ω=φ such that |KG(jφ)| = 1 and ∠(KG(jφ)) = -180°, then we know that s=jφ must be a pole.
This is incorrect, and is leading to a misunderstanding.
The poles of the system are the roots of the characteristic equation of the Open Loop transfer function $$KG(s)=0$$
the roots are of the form $$s=\sigma+j\omega$$
These are the poles and zeros that are analysed in the Bode Plot.
Once the loop is closed the poles move location to be the roots of the characteristic equation $$1+KG(s)=0$$ You can view loop compensation as a method of moving the open loop poles into more suitable (stable) locations in the closed loop. This can even by performed directly using pole placement.
However, Bode stability analysis is based on the Nyquist Stability Criterion. So the condition for oscillation in a negative feedback system is unity gain and 180 degree phase shift :$$KG(s) = -1$$
and therefore the Bode plot illustrates the "stability condition" by rearranging $$KG(s)+1=0$$
This happens to be the same equation as the characteristic equation of the Closed Loop. But it is a misunderstanding to see this as relating to Bode stability plots, which are an Open Loop analysis.
The Nyquist Stability Criterion also tells us that in general (but there are exceptions), a closed loop system is stable if the unity gain crossing of the magnitude plot occurs at a lower frequency than the -180 degree crossing of the phase plot.
Best Answer
The open loop Bode plot provides relative stability information on the closed loop system.
If an open loop can provide unity gain (or 0dB gain) in response to a sinusoid at a particular frequency where it also provides a -180deg phase shift then it can use the output sinusoid to replace the input sinusoid (the negative f/b provides the necessary additional -180deg phase shift). This is critical stability. Oscillation is self-sustaining. The loop does not see any difference between the externally applied sinusoid, and the one that it has created itself at the output terminal.
An open loop gain greater than 0dB at the frequency where the phase is -180deg will, clearly, give an unstable closed loop. And an open loop gain less than 0dB at that particular frequency will mean that the sinusoid gradually reduces in amplitude as it travels around the closed loop. That's a stable system.