It seems that the confusion is due to some unfortunate notation. As the OP states, Fano noise is due to the variance in photoelectron production per incident photon, and this should indeed be signal-dependent. However, the author also states that the total noise is given by:
$$\tag{1}
\sigma^2_\mathrm{TOTAL} = \sigma^2_\mathrm{READ} + \eta_i F_F + \eta_i S
$$
This equation is confusing, because the first and third term appear to be the familiar terms for read noise and Photon noise, but the second term, representing Fano noise, appears to have no dependance on the signal level. Matters are further confused because the notation used by the author is subtly different from what I (and I assume most others?) are familiar with.
First we will figure out this notation, then we will clarify the confusion about Fano noise, and finally we will see if we can make sense of equation 1. I'll be referring to the chapter of the text in question, "Photon Transfer" by James Janesick, which can be found for free here. I will explicitly state when I am referring to an equation from Janesick, while references to my own equations will simply be by number.
Notation
To be clear, I'm not saying this notation is bad. It's just different from what most people are used to, but similar enough to be confusing if you aren't careful.
$$ \begin{array} {l l}
\sigma^2 & \quad \textrm{Variance} \\
\eta_i & \quad \textrm{The expectation value for the numper of photoelectrons per incident photon} \\
F_F & \quad \textrm{Fano coefficient} \\
S & \quad \textrm{Expected Number of photoelectrons in an exposure} \\
N_P & \quad \textrm{Number of photons incident in a given exposure} \\
P_I & \quad \textrm{Expected Number of photons incident on the detector}\\
\end{array}
$$
Note the following relations as well:
$$\begin{array} {r c l}
P_I &=& \left<N_P\right> \\
S &=& \left< S_\textrm{PEAK} \right> \\
S_\textrm{PEAK} &=& \eta_i N_P \\
S &=& \eta_i P_I
\end{array}$$
where $\left<X\right>$ represents the expectation value of the random variable $X$. So remember that $S$ and $P_I$ are the mean values of their respective distributions, while $N_P$ and $S_\textrm{PEAK}$ are the specific values measured in a particular trial.
Another subtlety of this author's notation is that $\eta_i$ is not simply the quantum efficiency, and is not necessarily limited to [0, 1). In this author's notation, $\eta_i$ includes both of the quantities that I would refer to separately as the quantum efficiency $\eta$, and the gain $G$. So I may substitute $\eta_i \rightarrow \eta G$ where $\eta$ is a unitless and $0 \leq \eta \leq 1$.
As an example, the third term in equation 1 is supposed to represent the contribution to the variance of the photoelectron number due to the Poisson distributed statistics of the photon signal (i.e. photon noise). If we make all of my preferred substitutions we get:
$$ \begin{array}{r c l c r c l}
\sigma^2_\textrm{photon} &=& \eta_i S &\quad& S &\rightarrow& \eta_i P_I \\
\sigma^2_\textrm{photon} &=& \eta^2_i P_I &\quad& \eta_i &\rightarrow& \eta G \\
\sigma^2_\textrm{photon} &=& \eta^2 G^2 P_I
\end{array} $$
...which is the familiar expression for the variance contribution from photon noise is a system with gain. This also agrees with Janesick equations 3.5 and 3.6. So far, so good.
Fano Noise
For a moment let's ignore equation 1, and follow Janesick in section 3.3, where he introduces Fano noise. Janesick uses the example of a CCD or some other detector array, so we may, interchangeably, speak of the signal from a single "trial," or from one pixel, or from a single exposure. Furthermore, each pixel in the array will see a photon source with the same $P_I$, while the number of photons that hit each pixel, $N_P$, will generally not be exactly $P_I$, but will follow a Poisson distribution with $P_I$ as its mean.
In Janesick equation 3.7, he gives the variance in electron-hole generation as:
$$\tag{2}
\sigma^2_\textrm{FN} = F_F \eta_i
$$
where $F_F$ is a constant, $0\leq F_F \leq 1$ describing the strength of the noise. Looking at this equation, we can be confident that we are considering the electron-hole production from a single photon interaction. Why? Well, consider the case where $F_F = 0$. If this expression is for a single photon interaction, then the $F_F = 0$ case corresponds to the situation where each photon produces exactly the same number of photoelectrons, and there is no extra noise term associated with the electron-hole production process. Now consider the $F_F = 1$ case. This, according to the text preceeding the equation in Janesick, corresponds to the electron-hole production process being purely Poissonian and indeed the equation would reduce to
$$ \sigma^2_\textrm{FN} = \eta_i $$
which simply tells us that the variance from the electron-hole generation process would be equal to the expected number of photoelectrons from one photon($\eta_i N_P$ where $N_P = 1$). In other words the variance would equal the mean, as expected for a Poisson process.
So now we understand Fano noise for a single photon interaction, and we see that $F_F$ describes the degree to which this noise is less noisy than a simple Poisson process.
Moving on, Janesick goes back to the CCD array example, and plots a histogram of the charge measured on each pixel after exposing the array to an average incident signal of 3 photons per pixel, with $\eta_i = 10e^- \textrm{/photon}$. So we are looking at the result of many trials with $P_I=3 \textrm{photons}$ and $S = 30e^-$, but there are peaks around $(10, 20, 30, \ldots) e^-$ per pixel because $N_P$ is Poisson distributed. Each peak is broadened by read noise, but we also see that the peaks at larger $N_P$ are wider. Why? Janesick equation 3.9 tells us:
$$ \sigma^2_\textrm{FN,PEAK} = N_P F_F \eta_i $$
In other words, Fano noise varies as a function of signal, but not the mean signal $P_I$. Rather, it varies with the number of incident photons incident in a single trial.
Now things become confusing. Janesick tells us that to combine shot noise and Fano noise, we add the variances in the normal way:
$$ \sigma^2_\textrm{SHOT+FN} = \sigma^2_\textrm{SHOT} + \sigma^2_\textrm{FN} $$
No surprise there, but we are wondering what expression will give us $\sigma^2_\textrm{FN}$? Shot noise, as we know, varies with the signal $P_I$ or $S$, but we have not yet seen how to extend the $N_P$ dependance of Fano noise to an expression containing the mean signal level. At this point we understand Fano noise as a contribution to the variance in photoelectron number from pixels with a given number of incident photons, but this number itself is Poisson distributed. Janesick substitutes equation 2 (his equation 3.7) to state that
$$\tag{3}
\sigma^2_\textrm{SHOT+FN} = \eta_i \left(S + F_F \right) $$
..but he does not explicitly justify this substitution.
The Confusion, and Resolution
It is this final substitution which leads to the form we see in equation 1, and the confusion can be traced to a lack of clarity on how that substitution is justified. Possibly there is some theorem which states that variance of a process in which the variances of each trial are themselves distributed with a given variance can be added in such a way as to arrive at equation 3, but I am unaware of such a theorem.
I suspect that the correct resolution to this confusion is two-fold: First, the noise term described in Janesick equation 3.7 is purely an empirical description, so one could assume that the form of equation 1 is simply an empirical estimate of noise in a detector where Fano effects are non-negligible.
Second, it is only for small signals where Fano noise will dominate, so when considering Fano noise we can take $N_P \approx 1$. Thus we can justify the substitution made by Janesick. This assumption would seem appropriate, as the signal to noise ratio (SNR) due to photon noise will scale with $\sqrt{P_I}$, while the SNR for Fano noise will scale like $\sqrt{\frac{N_P \eta_i}{F_F}}$, so for Fano noise to dominate over photon noise would imply high gain and low photon flux. In other words, Fano noise is associated with single photon counting detectors.
The practice for calling signals digital or analog in communications systems is that analog signals carry information in continuous or semi-continuous changes in the signal, whereas digital ones carry information in discrete changes. Discrete and digital are used interchangeably. See for instance Stallings basic books, or see modern and excellent books in communications like Proakis and Salehi Communications Systems Engineering. Wikipedia may also be ok, but they're never really correct (I teach a course on it and have to always warn the students). Since the information carried is the relevant factor, the provisioning of information to some electromagnetic (or other) wave (also called the carrier wave, or carrier) is called modulation, or sometimes more generally encoding.
Wiki is at https://en.wikipedia.org/wiki/Digital_signal. No guarantees.
After the modulation everything, carrier and modulation, ie, the full signal, is usually upconverted to the frequency at which it is meant to be transmitted, sent tot he antenna, and radiated away. Yes, of course, the actuial physical wave sent out, with the carrier imparted with the modulation (ie, the changes), is analog, in the sense that even the discrete changes take a small amount of time, and the power being sent does have some, typically very small, power in the time during which is 'discretely' changes.
Thus an FM modulated signal is analog, also AM and PM, where the changes are continuous. The digital (discrete) equivalents are FSK, ASK and PSK, where the freq, amplitude or phase changes quickly, called discontinuously (more on this in next paragraph, it always takes a bit of time as stated above). In digital signaling there is a lot lot lot more control and precision one can exercise over the signal features, and thus more information. Digital circuits also make it much cheaper.
As for those discontinuities, as you probably know from Fourier transforms and from quantum mechanics, any fast change in the time domain of the signal, called the waveform to emphasize it's changing shape, will mean the time uncertainty of the change is small, and thus the energy will be spread over a larger range, creating what is interference in nearby frequency bands. Thus part of the trick (now a well understood set of tradeoffs) is to have some shaping of the transition for the optimal tradeoff desired. In modern LTE 4G wireless comms the carriers are separated via an inverse Fourier transform so that the interference with adjacent frequencies is theoretically zero, at the specific transition time, but again, still some design tradeoffs.
So, yes if you are a physicist or an RF engineer or technician you might say all signals are analog. But the practice in communications engineering (systems and design engineers, practitioners and technicians) is to call those with very rapid transitions, digital.
Best Answer
Treating the signals as time series: If the first signal $S_1$ has a noise component $N_1$ added to it, then the noisy signal is $S_1+N_1$, similarly the second signal is $S_2+N_2$, so the difference signal would be $(S_1+N_1)-(S_2+N_2)$ and its signal to noise ratio would be $\langle(S_1-S_2)^2\rangle\over\langle(N_1-N_2)^2\rangle$
If the signals are uncorrelated $\langle(S_1-S_2)^2\rangle$ is just $\langle S_1^2\rangle+\langle S_1^2\rangle$. If the signals are correlated, you will have to estimate the covariance $\langle S_1S_2\rangle$ to compute $\langle S_1^2\rangle+\langle S_1^2\rangle-2\langle S_1S_2\rangle$
Ditto for the noise.