First, the finite sample variance that you have above is called the Neyman conservative variance. This result requires a little trick about how we write the estimator and an assumption of a random sample from an infinite population.
Define, $K=\frac{n_1}{n}$. Now consider,
$$\hat\tau = \bar{Y_i(1)}-\bar{Y_i(0)}=\frac{1}{n_1}\sum_{i=1}Z_iY_i - \frac{1}{n_0}\sum_i (1-Z_i)Y_i \\= \frac{1}{n}\sum_{i=1}\frac{Z_iY_i}{K} - \frac{1}{n}\sum_i \frac{(1-Z_i)Y_i}{1-K}\\=\frac{1}{n}\sum_i(\frac{Z_iY_i}{K}-\frac{(1-Z_i)Y_i}{1-K})=\frac{1}{n}\sum_i(\frac{Y_i(1)}{K}-\frac{Y_i(0)}{1-K})$$
Define the population average treatment effect or PATE by,
$$\tau = \mathbb{E}[Y_i(1)-Y_i(0)]$$
Then, since we have that these are iid sequences by the WLLN,
$$\hat\tau \overset{p}{\to} \tau$$
Notice that $K\overset{p}{\to}\Pr[Z_i=1]$. So by Slutzky's theorem and the CLT we have,
$$Avar(\hat\tau)= \frac{Var(Y_i(1))}{\Pr[Z_i=1]}+\frac{Var(Y_i(0))}{\Pr[Z_i=0]}$$
where $Avar(\cdot)$ refers to the asymptotic variance or the variance of the limiting distribution of the estimator.
Yielding,
$$\sqrt{n}(\hat\tau - \tau)\overset{d}{\to}N(0,\frac{Var(Y_i(1))}{\Pr[Z_i=1]}+\frac{Var(Y_i(0))}{\Pr[Z_i=0]})$$
As desired.
Addendum:
I glossed over this in the main answer but on second thought I think it is useful to note. It is easy to see that the variance of the parameter would be given by,
$$Var(\hat\tau) = \frac{n}{n_1} Var(Y_i(1)) + \frac{n}{n_0} Var(Y_i(0)) - Cov(Y_i(1),Y_i(0))$$
Of course, this is not identified within the sample. However under asymptotic inference, given true random sampling from the population, the covariance term will be $0$ because control and treatment are sampled independently. Thus, yielding the asymptotic variance that we see above.
For anyone interested in results under more general assumptions, I would recommend this paper by Li et al..