The Nelson-Aalen estimator is $ \hat{\Lambda}(t) = \sum \frac {d_j}{r_j}.$
I specifically want to use Greenwood's Formula such that
$$ Var(\hat{S}(t)) = \hat{S}(t)^2\sum_j \frac{d_j}{r_j(r_j-d_j)}$$
Now, using the NA estimator, we can use some identities to show
$$ \hat{S}(t) = e^{-\hat{\Lambda}(t)}$$
Then
$$ Var(\hat{S}(t)) = e^{-\hat{\Lambda}(t)} \sum_j \frac{d_j}{r_j(r_j-d_j)} $$
I want to find $Var(\hat{\Lambda}(t))$ from this, but $log(Var(\hat{S}(t))\neq Var(log(\hat{S}(t)).$ So what do I do?
It has been years since I've touched this material, but I'll give it a shot. This procedure is outlined in p. 233 of Loss Models: From Data to Decisions, 4th ed., by Klugman, Panjer, and Willmot.
Assume the number of deaths at time $t_i$ has a Poisson distribution with parameter $r_ih(t_i)$, where $h$ is the hazard rate. Its variance is thus $r_ih(t_i)$, with $h(t_i) \approx d_i / r_i$, so the approximate variance is thus $r_i(d_i/r_i) = d_i$.
Assuming independence, $$\widehat{\text{Var}}(\widehat{\Lambda}(t)) = \widehat{\text{Var}}\left(\sum \dfrac{d_j}{r_j} \right) = \sum\dfrac{\widehat{\text{Var}}(d_j)}{r_j^2} \approx \sum\dfrac{d_j}{r_j^2}\text{.}$$