Discrete Time Fourier Transform (DTFT) for an unstable system (Ideal Low Pass Filter)
It is generally not true that the relation $Y(\omega)=X(\omega)H(\omega)$ is independent of the system's stability. For systems with a rational transfer function (i.e., systems that can be described by differential or difference equations), the frequency response, which is the Fourier transform of the impulse response, only exists for stable systems. For marginally stable systems with poles on the imaginary axis (for continuous-time systems), or on the unit circle (for discrete-time systems), the Fourier transform exists if we allow Dirac delta impulses in the expression for the Fourier transform. So for marginally stable systems you can see that the output can become unbounded, even for bounded input signals, due to the Dirac delta impulses in their frequency response. For unstable causal systems with transfer function poles in the right half plane, the frequency response doesn't exist.

For systems for which there exists no transfer function, but which do have a frequency response, such as ideal frequency-selective filters (low pass, high pass, etc.), the relation $Y(\omega)=X(\omega)H(\omega)$ holds, regardless of stability. Ideal frequency-selective filters are unstable (and non-causal), but their frequency response $H(\omega)$ is finite. This means that the frequency response of the bounded input signal which generates an unbounded output (if it exists) must be unbounded for at least one value of $\omega$, and the inverse Fourier transform of $X(\omega)H(\omega)$ must result in an unbounded time-domain signal. Also note that there is no guarantee that the Fourier transform $X(\omega)$ of the bounded input signal that generates an unbounded output signal even exists. And if it exists, it's generally not immediately clear from the expression $X(\omega)H(\omega)$ that the corresponding inverse transform is unbounded.

I think it is a misunderstanding that the relation $Y(\omega)=X(\omega)H(\omega)$ somehow implies that a bounded input must result in a bounded output. It's just not so easy to see whether the output is unbounded because one would have to compute

$$\frac{1}{2\pi}\int_{-\pi}^{\pi}X(\omega)H(\omega)e^{jn\omega}d\omega$$

and check for each value of $n$ if the expression becomes unbounded. Direct time domain analysis is more straightforward in this case.