This is not really an answer to your question because I'm not sure there is a useful general answer. You say, "Calculating statistical power seems to be straightforward for paired/non-paired t-tests." Let's start there.
Background on the power of t tests. For example suppose we want to find power for a two-sample pooled t test. What do you need to know in order to get the sample size $n = n_1 = n_2$ in each group in order
to detect a difference $\Delta = \mu_1 - \mu_2$ (or larger) with $90\%$ power in a one-sided test with significance level $\alpha = $5%.$
In addition to $\alpha,$ $\Delta$ and the desired power, you need to know the common population standard deviation $\sigma_1 = \sigma_2 = \sigma.$ More specifically, you need
to know $\Delta/\sigma.$ Suppose, you want $\Delta/\sigma = 0.5.$ Then you are
ready to use the 'power and sample size' procedure in some statistical software to
find the required $n = 70.$ Results from Minitab are shown below, based on the noncentral t distribution. The power curve
shows the power for other values of $\Delta$ when other quantities are the same.
Power and Sample Size
2-Sample t Test
Testing mean 1 = mean 2 (versus >)
Calculating power for mean 1 = mean 2 + difference
α = 0.05 Assumed standard deviation = 2
Sample Target
Difference Size Power Actual Power
1 70 0.9 0.902966
The sample size is for each group.
A similar computation for the Welch one-sided, two-sample t test is a little messier
because the Welch test does not require $\sigma_1 = \sigma_2$ and compensates for
unequal group variances by decreasing the degrees of freedom based on the sample standard deviations and sample sizes. For a specific case, I can use simulation
to check whether $n = 70$ subjects in each group is still sufficient for $90\%$ power
if $\sigma_1 = 1.5, \sigma_2 = 2.5.$ The answer is Almost, the power drops to a little over $89%.
set.seed(2010)
pv = replicate(10^5, t.test(rnorm(70, 50, 1.5),
rnorm(70, 51, 2.5), alt="less")$p.val)
mean(pv <= .05)
[1] 0.88639 # Rejection probability = power
Specific information required for power of Wilcoxon RS test. Now for nonparametric two-sample Wilcoxon rank sum tests. Somehow we would need
to know the effect size $\Delta$ and the variability of the data in order to get
the required sample size for a test at level $\alpha.$ If you could be sufficiently
specific about the two populations being compared, you could use simulation
to find the power of the Wilcoxon test for a given number $n$ of subjects in each
group. It is not enough just to know that the population distributions are not normal.
If you had evidence that the two populations are $\mathsf{Gamma}(\mathrm{shape}=5,
\mathrm{rate}=0.13)$ and $\mathsf{Gamma}(5, 0.1),$ which are of roughly the
same shape, you could use a Wilcoxon RS test to see if the populations have
different medians. For $n = 70$ subjects in each group the power would be about $95\%.$
set.seed(2010)
pv = replicate(10^5, wilcox.test(rgamma(70, 5, 0.13),
rgamma(70, 5, 0.1), alt="less")$p.val)
mean(pv <= .05)
[1] 0.94822
The two medians are about $\eta_1 = 35.9, \eta_2 = 46.7.$
qgamma(.5, 5, 0.13); qgamma(.5, 5, 0.1)
[1] 35.93007
[1] 46.70909
However, if we have such specific information about the two population
distributions, we might be able to find an exact test, using the mathematical
forms of the density functions, which could be more powerful than the Wilcoxon SR test.
Perhaps the main difficulty finding the power of a nonparametric test is that
we do not usually think about the specific form of the populations being compared.
The term 'nonparametric' often means 'not necessarily just for normal populations'. It does not
forbid application to distributions that have density functions involving parameters.
