The bilinear transform compresses the analog frequency in the s-domain associated with $f=\infty$ to the point $z=-1$ in the z-domain (which is where the frequency is $f_s/2$ where $f_s$ is the sampling rate. Consider what the magnitude of your notch filter is at $f=\infty$: It has no effect on the magnitude or phase of your signal; there is no filtering impact there whatsoever.
Compare in contrast to this your low pass filter (and how much of your "desired" shape is already beyond the Nyquist boundary which no pre-warping would be able to reconstruct). As you approach the Nyquist boundary the ability to compensate with pre-warping becomes increasingly difficult (you will see the same with the notch if it is brought close enough such that there is continued attenuation beyond the boundary).
The degree of this frequency warping and how it pronounced is depicted in the chart example that I pasted below (not the OP's sampling rates) which shows which frequency on the vertical axis for a given sampling rate of 20 rad/sec (125.6 Hz) would map to the digital frequency domain extending from $\pi$ to $\pi$ rad/sample.
With prewarping, especially at high frequencies, it is very difficult to match an analog shape over a wider frequency range given this mapping is according to the following transcendental equation (would require Taylor series expansion out to an increasingly higher order as we near the Nyquist boundary to match effectively, but is easily matched with the first or second terms only at the lower frequencies where the slope becomes approximately linear):
$$\omega_a = \frac{2}{T}tan(\omega_d T/2)$$
(as referenced here: https://en.wikipedia.org/wiki/Bilinear_transform)
But for a single point, or narrow range as typical for a notch filter, this repositioning of a single notch zero would be trivial to do and with high accuracy for the repositioned notch location (but distortion right at the Nyquist boundary if the target filter has residual insertion loss beyond Nyquist).
