Previous answers (from @Hilmar, @RickLyons, and @hotpaw2) are very informative. Hope I can also provide some.
First of all, there are more than one way to construct an complex signal based on a real one. Think about
$x(t)=\cos(2\pi f_1 t)\cos(2\pi f_2 t)$, where $f_1<f_2$,
one can define either
$z(t)=\cos(2\pi f_1 t)e^{j2\pi f_2 t}$, or $z(t)=\cos(2\pi f_2 t)e^{j2\pi f_1 t}$
However, Hilbert transform only generates one analytic signal (the first one). According to Bedrosian's product theorem, for real signal $x(t)=a(t)\cos\phi(t)$, its Hilbert transform
$H\{a(t)\cos\phi(t)\}=a(t)H\{\cos\phi(t)\}=a(t)\sin\phi(t)$
if $|F\{a(t)\}|=|A(\omega)|>0$ only for $\omega<\omega_{critical}$,
while $|F\{\cos\phi(t)\}|>0$ only for $\omega>\omega_{critical}$.
In this case, $a_z(t)=|x(t)+jH\{x(t)\}|=a(t)$ will be the desirable envelope. Otherwise, $a_z(t)\ne a(t)$ and might be hard to interpret. Please refer to [2] for more details.
[I'm not quite sure about my following reasoning...]
The analytic signal associated with any real bandpass ($\omega_a<\omega<\omega_b$) signal can be expressed as
$z(t)=A(t)e^{j(\omega_a+\phi(t))}=A(t)e^{j\phi(t)}e^{j\omega_a}=z_{base}(t)e^{j\omega_a}$,
where the spectrum of $z_{base}(t)$ and thus $A(t)$ is upper bounded by $\Delta\omega=\omega_b-\omega_a$,
while the spectrum of $e^{j\phi(t)}e^{j\omega_a}$ is lower bounded by $\omega_a$
If $\omega_b<2\omega_a$ (or $\Delta\omega<\omega_a$), then $H\{A(t)\cos(\omega_a+\phi(t))\}=A(t)\sin(\omega_a+\phi(t))$,
and Hilbert transform will provide reasonable envelope.
I think this is why to bandpass a broadband signal (e.g., music audio) in a series of octaves before applying Hilbert transform results in reasonable outputs.
[2]: Boashash, B. (1992). Estimating and interpreting the instantaneous frequency of a signal. I. Fundamentals. Proceedings of the IEEE, 80(4).