Male-female height difference estimation from thresholds Introduction. Assume that two populations $A$ and $B$ are distributed with normal distributions $N(\mu_A, \sigma_A^2)$ and $N(\mu_B, \sigma_B^2)$.
This is a general problem, but as an example, I will consider height distributions of men and women. Assume that the heights of men and women are normally distributed with the following means and standard deviations:
$$\begin{align} & \textbf{Mean} & \textbf{St.d.}\\
\textbf{Female}\quad & 161\,\text{cm} & 7.1\,\text{cm}\\
\textbf{Male}\quad& 175\, \text{cm} & 8.5\,\text{cm} \end{align}$$
Importantly, I don't know these parameter values, but I wish to infer/estimate them (Read on). Instead, what I have is a table of the proportion of men and women exceeding some (unknown) threshold. For example, I have manually created some thresholds. The following table contains the proportion of women and men, respectively, who exceed a certain threshold of height:
$$\begin{align} \textbf{Threshold} &\quad \textbf{Female} & \textbf{Male}\\
\geq158\,\text{cm} &\quad0.6637 & 0.9772\\
\geq 162\,\text{cm} & \quad 0.4440 & 0.9369\\
\geq 166\,\text{cm} & \quad 0.2406 & 0.8552 \\
\geq 170\,\text{cm} & \quad 0.1025 & 0.7218\\
\geq 174\,\text{cm} & \quad 0.0336 & 0.5468\\
\geq 178\,\text{cm} & \quad 0.0083 & 0.3621\end{align}$$
I only have the actual proportion values (0.6636, 0.9772, ... etc) but I don't have the threshold values (158 cm, 162 cm, etc) as they are unknown.
Question. Given such a threshold table, how do I find the group mean difference (in SD units) and variance ratio? (Assuming that they are normally distributed)
Results.
Given I have manually constructed the data, I can say what the results should approximately be. If we standardize the female distribution, then the mean difference is
$$\Delta := \frac{\mu_M-\mu_F}{\sigma_F} = \frac{175-161}{7.1} \approx 1.972 \quad\text{female standard deviation units} $$
The variance ratio is 
$$\rho^2 := \frac{\sigma_M^2}{\sigma_F^2} = \frac{8.5^2}{7.1^2} \approx 1.433$$
Which are the two values that I'm looking for (mean difference and variance ratio).

Thoughts. My thinking is to first reduce the problem from four unknown variables ($\mu_A$, $\mu_B$, $\sigma_A^2$, $\sigma_B^2$) to a problem of two unknowns. This can be done by standardizing one group to be distributed with mean 0 and variance 1. Then I am simply looking for the mean difference (say, $\Delta$) and the variance ratio.
Then I probably have to calculate some density function or cumulative density function for the difference, and make a least squares optimization problem with the table (There can be noise in the data, hence I am thinking a least squares minimization). But I'm unsure exactly how.
 A: (I created a solution to the problem myself, so I give it as an answer here for anyone who is interested.)
Method
First, we have the data that we wish to estimate group differences from. These indicate the proportions of women and men, respectively, that exceed a certain threshold. For example, when $66.36\%$ of women are taller than some threshold, then $97.72\%$ of men are taller than that same threshold. Solution code is given in R. Here is the data from the original question:
P_F = c(0.6636, 0.4440, 0.2406, 0.1025, 0.0336, 0.0083)
P_M = c(0.9772, 0.9369, 0.8552, 0.7218, 0.5468, 0.3621)

First, we standardize the female height distribution. They have mean 0 and standard deviation 1. From this, we can calculate the underlying threshold values $t_1, t_2, ..., t_n$ (where $n = 6$ is the number of thresholds, i.e. the length of one of the vectors above). We do this by solving for $t_i$ such that
$$\int_{t_1}^\infty f(t;\mu = 0,\sigma = 1)\,\mathrm{d}t = 0.6636$$
where $f(t; \mu, \sigma)$ is the normal density function with mean $\mu$ and standard deviation $\sigma$. We do this for all values in P_F (this was just the first one). Here is the code:
f0 = function(x){ # Standard normal density function
     1/sqrt(2*pi)*exp(-x^2/2)
}
# Find threshold values
thresholds = rep(0, length(P_F))
for(i in 1:length(P_F)){
     F0_temp = function(t){
     -P_F[i] + integrate(f0,t,Inf)$value
     }
thresholds[i] = uniroot(F0_temp, lower = -10, upper = 10)$root
}

Once we have the threshold values $t_1, t_2, ..., t_n$ we want to find values $\Delta$ and $\rho$ such that $f(t,\Delta,\rho)$ is the best fit to the data P_M. First we create the function that we wish to fit
# Create function to fit
F_fit = function(t,Delta, rho){
     f0_full = function(x, mu = Delta, sigma = rho){
          1/sqrt(2*pi*sigma^2)*exp(-(x-mu)^2/(2*sigma^2))
     }
     outp = rep(0, length(t))
     for(i in 1:length(outp)){
          outp[i] = integrate(f0_full,t[i],Inf)$value
     }
     return(outp)
}

Next we find the best fit
# Fit using nonlinear least squares
fitLS = nls(P_M ~ F_fit(thresholds, Delta, rho), start = list(Delta = 1.6, rho = 1.2))

The found parameters are then given by
# Store parameters
Delta = unname(coef(fitLS)[1]) # Mean difference (in SD units)
rho = unname(coef(fitLS)[2]) # Standard deviation ratio

Results
We can check if the results come close to the expected values (correct values), with $\Delta$ being the mean difference and $\rho^2$ being the variance ratio.
> Delta
[1] 1.972079
> rho^2
[1] 1.434806

As we can see, they are very close to the expected values $1.972$ and $1.433$ (see original question). The method seems to perform decently (at least in this case).
Plot
I now create a plot with the following code
# Create plot
linspac = seq(0,1, length.out = 100) # Line of equality
plot(linspac, linspac, type = "l", main = "Best fit", xlab = "Proportion of females exceeding threshold", ylab = "Proportion of males exceeding threshold")
points(P_F, P_M, col = "red") # Data points

tvals = seq(-5,5,length.out = 100)
xvals = rep(0,length(tvals))
yvals = rep(0,length(tvals))

for(i in 1:length(tvals)){ # Create curve
     xvals[i] = F_fit(tvals[i], 0, 1)
     yvals[i] = F_fit(tvals[i], Delta, rho)
}

lines(xvals, yvals, col = "blue") # Plot curve

Here is the plot from the code (annotated in paint):

