A general result will be difficult. Sometimes, eg if $p_1$ and $p_2$ were the same in your example and you wanted to estimate that common $p$, the bound will be lower than the separate bounds. In that setting I believe the bound on the precision of the common $\hat p$ will be the sum of the bounds on precisions of the individual $\hat p_i$. That is $$\mathrm{var}^{-1}[\hat p]\leq \mathrm{var}^{-1}[\hat p_1]+\mathrm{var}^{-1}[\hat p_2]\leq n_1I_1+n_2I_2$$ where $I_i$ is the per-observation Fisher information for $p$ in sample $i$.

In your example, though, you can't get a bound for $d$ in the combined data from bounds for $d$ in the two separate samples because $d$ isn't identifiable from a single sample. You'll need something that considers them together. (well, for $p_1-p_2$ you could work it out, but general functions would be hard)

An asymptotic version would be more tractable. Write the data as $(i, S_{i,n})$ for $i=1,2$, $n=1,\dots,n_i$. It doesn't matter asymptotically whether $n_i$ are fixed or whether each observation is independently assigned to a set, so we can treat $(I, S_{I,n})$ as iid again, where $I$ is Bernoulli($q$) for some $q$. Now, $d$ is some differentiable function of $(q,p_1,p_2)$ and we can write down the Fisher information about $d$ and invoke the usual efficiency results -- one of the Convolution theorems or the local asymptotic minimax theorem. It doesn't matter whether $\hat p_i$ or $\hat d$ are finite-sample unbiased, which is another simplification for asymptotics, since for non-linear functions $d$ there will often be no unbiased estimators.

I don't think there's a shortcut to working out the Fisher information for $d$, which is the key step and which handles the distinction between $(\hat p_1,\hat p_2)\mapsto \hat p$ and $(\hat p_1,\hat p_2)\mapsto \hat p1-\hat p_2)$.

A general result will be difficult. Sometimes, eg if $p_1$ and $p_2$ were the same in your example and you wanted to estimate that common $p$, the bound will be lower than the separate bounds. In that setting I believe the bound on the precision of the common $\hat p$ will be the sum of the bounds on precisions of the individual $\hat p_i$. That is $$\mathrm{var}^{-1}[\hat p]\leq \mathrm{var}^{-1}[\hat p_1]+\mathrm{var}^{-1}[\hat p_2]\leq n_1I_1+n_2I_2$$ where $I_i$ is the per-observation Fisher information for $p$ in sample $i$.

In your example, though, you can't get a bound for $d$ in the combined data from bounds for $d$ in the two separate samples because $d$ isn't identifiable from a single sample. You'll need something that considers them together. (well, for $p_1-p_2$ you could work it out, but general functions would be hard)

An asymptotic version would be more tractable. Write the data as $(i, S_{i,n})$ for $i=1,2$, $n=1,\dots,n_i$. It doesn't matter asymptotically whether $n_i$ are fixed or whether each observation is independently assigned to a set, so we can treat $(I, S_{I,n})$ as iid again, where $I$ is Bernoulli($q$) for some $q$. Now, $d$ is some differentiable function of $(q,p_1,p_2)$ and we can write down the Fisher information about $d$ and invoke the usual efficiency results -- one of the Convolution theorems or the local asymptotic minimax theorem. It doesn't matter whether $\hat p_i$ or $\hat d$ are finite-sample unbiased, which is another simplification for asymptotics, since for non-linear functions $d$ there will often be no unbiased estimators.

I don't think there's a shortcut to working out the Fisher information for $d$, which is the key step and which handles the distinction between $(\hat p_1,\hat p_2)\mapsto \hat p$ and $(\hat p_1,\hat p_2)\mapsto (\hat p_1-\hat p_2)$.