It is not valid in general because the deduction of the distribution of the t-statistic depends on the assumption of normality. In addtion, the interpretration of testing the hypothesis of equality of means in the half-normal case is not clear because the mean depends on the parameters mu and sigma. Therefore if the variance of the populations differs (even slightly), you will reject the hypothesis more than you should even if the parameters mu1=mu2 (this is, you are not getting the desired Type I error level). Take a look at this code so you can verify it
rm(list=ls())
power = function(n,ns,mu1,mu2,s1,s2){
count = rep(0,n)
for(i in 1:n){
x = mu1 + abs(rnorm(ns,0,s1))
y = mu2 + abs(rnorm(ns,0,s2))
if(t.test(x,y,paired=TRUE)$p.value>0.05) count[i] = 1
}
return(1-mean(count))
}
power(10000,30,1,1,1,1.15)
power(10000,100,1,1,1,1.15)
power(10000,30,1,1.1,1,1)
Perhaps it would be helpful to plot the profile likelihood of the parameters of each population and take a look at the similarities between them. Also you could use a likelihood ratio test instead.
I hope this helps.
Best wishes.