Comment. It seems to me you may not be on a useful track. Maybe this will help.
Let $\bar X$ be the sample mean, $S^2$ be the sample variance, and
$\tau = \mu/\sigma.$
Then by the method of moments it might be reasonable to look at
$\hat \tau = \bar X/S$ as an estimate of $\tau$ based on sufficient
statistics. One would not necessarily expect it to be an unbiased
estimator (perhaps asymptotically unbiased), but might check to see how biased it is. Because $S$ is known to be slightly negatively biased for $\sigma$
it is not surprising that $\hat \tau$ is slightly positively biased for $\tau.$
I do not want to claim this is the path you should take, but I think it
may make more sense than what you have suggested. (Your denominator does
not seem feasible because your $S^2$ and $\bar X$ have different dimensionalities.)
With a quick simulation of a million samples of size $n = 10$ from
$Norm(\mu = 100, \sigma=20)$ (so that $\tau = 5$), in R statistical software, we can investigate this idea. At the end of my simulation
I have shown a modification of your estimator that may have promise.
m = 10^6; mu = 100; sg = 20; n = 10
x = rnorm(m*n, mu, sg); DTA = matrix(x, nrow=m) # each row a sample
a = rowMeans(DTA); s = apply(DTA, 1, sd); tau.hat = a/s # m-vectors
mean(a); sd(a); mean(s); sd(s)
## 99.99136 # exact is 100
## 19.45392 # as known, S slightly negatively biased for sg
## 5.469827 # not surprisingly, somewhat positively biased for tau
# investigating your estimator
SS = rowMeans(DTA^2) # your 'S^2'
your.est = a^2/(SS - a)
## 0.9745633 # nowhere near tau
alt.est = sqrt(a^2/(SS - a^2)) # possible candidate
## 5.765703 # biased, but in the 'ballpark'