This question essentially asks whether the associativity of convolutions can be extended to cover the Hilbert transform, even when one does not know a priori that the Hilbert transform can be applied.
Let $\,g \in L^1(\mathbb{R}) \cap L^2(\mathbb{R})\,$ be a function whose Hilbert transform $h$ also belongs to $L^1(\mathbb{R})$.
Is it the case that for every $f \in L^\infty(\mathbb{R})$, for almost all $t \in \mathbb{R}$, $$ \frac{1}{\pi}\int_{(A,-\varepsilon) \cup (\varepsilon,B)} \!\!\frac{(g \ast f)(\tau)}{t-\tau} \, d\tau \ \to \ (h\ast f)(t) \hspace{5mm} \textrm{as } \ (\varepsilon,A,B) \to (0^+,-\infty,\infty) \, ?$$
If not: is it at least the case that for every $f,\tilde{f} \in L^\infty(\mathbb{R})$ and every $\tilde{g} \in L^1(\mathbb{R}) \cap L^2(\mathbb{R})$ whose Hilbert transform $\tilde{h}$ belongs to $L^1(\mathbb{R})$, if $\,g \ast f \overset{\textrm{a.e.}}{=}\tilde{g}\ast\tilde{f}\,$ then $\,h \ast f \overset{\textrm{a.e.}}{=}\tilde{h}\ast\tilde{f}$?
Physical motivation: The Hilbert transform is involved in the standard definition of the "instantaneous phase" of a signal. Now as I understand it, one cannot necessarily take the Hilbert transform of a general $L^\infty$ function (e.g. what would the Hilbert transform of $\mathrm{sgn}(x)$ be?). However, if one is looking for the "instantaneous phase", this suggests that the signal is being regarded as some kind of "simple" oscillator; in practice, one may wish to extract an individual oscillatory component from within some signal $f$ and then examine the instantaneous phase of that component. In this case, if the frequency range of the oscillatory component is known, one may be able to extract the component by filtering the pre-recorded signal $f$ through a zero-phase-distortion bandpass filter $f \mapsto g\ast f $. Here, $g$ will typically be the inverse Fourier transform of a rational function $\hat{g}$ with the following (among other) properties: there are no real poles, the order of the denominator is at least two higher than the order of the numerator, and $0$ is at least a double zero of $\hat{g}$. (This implies in particular that the Hilbert transform of $g$ is continuous and decays faster than $1/x^2$.) Of course, one can only record a finite-length extract from the behaviour of the system under investigation, and so my question is essentially about the physical meaningfulness of the above procedure. (Thus, in a sense, my question is a follow-up to https://mathoverflow.net/questions/293137/how-far-can-the-domain-of-definition-of-multiplier-operators-be-extended.)
The space of functions $g$ such that their Hilbert transform is in $L^1(\mathbb{R})$ is known to be the (real) Hardy space $H^1(\mathbb{R})$, see [Stein: Chapters III and IV]. Probably your associativity property follows for every $g \in H^1(\mathbb{R})$ just by proving it a priori for $g$ in the Schwartz class, taking limits in $H^1(\mathbb{R})$ using density and that $\| H(g) \|_{L^1} \sim \| g \|_{H^1}$.
[Stein]: Stein, Elias M., Harmonic analysis: Real-variable methods, orthogonality, and oscillatory integrals. With the assistance of Timothy S. Murphy, Princeton Mathematical Series. 43. Princeton, NJ: Princeton University Press. xiii, 695 pp. (1993). ZBL0821.42001.