Assuming that your Taylor expansion/Poisson arguments are enough to handle the case $np$ bounded, I believe the case $np \rightarrow \infty$ can be taken care of using the law of large numbers/Chebyshev. We have $$I_n(p) = E( H(p) - H(X))$$ where $X$ is $\frac{1}{n}Bi(n,p)$. Assume WLOG $p \leq \frac{1}{2}$, and take arbitrary $t<p$. We can bound $H(p)-H(X)$ above by $H(p)-H(t)$ for $t \leq X \leq 1-t$, and above by $1$ in general. This means $$I_n(p) \leq [H(p)-H(t)] + P(X<t) + P(X>1-t)$$

If, say, $t=p-(np)^{-1/3}$ and $np \rightarrow \infty$, then by Chebyshev's inequality the latter two terms are both at most $(np)^{-1/3}$, while the first term goes to $0$ because $H$ is continuous.

EDIT: For reasons explained by TMM in the comment below, these bounds aren't in and of themselves good enough. I'm leaving them here because the comment is valuable, but this doesn't work as an answer.

Assuming that your Taylor expansion/Poisson arguments are enough to handle the case $np$ bounded, I believe the case $np \rightarrow \infty$ can be taken care of using the law of large numbers/Chebyshev. We have $$I_n(p) = E( H(p) - H(X))$$ where $X$ is $\frac{1}{n}Bi(n,p)$. Assume WLOG $p \leq \frac{1}{2}$, and take arbitrary $t<p$. We can bound $H(p)-H(X)$ above by $H(p)-H(t)$ for $t \leq X \leq 1-t$, and above by $1$ in general. This means $$I_n(p) \leq [H(p)-H(t)] + P(X<t) + P(X>1-t)$$

If, say, $t=p-(np)^{-1/3}$ and $np \rightarrow \infty$, then by Chebyshev's inequality the latter two terms are both at most $(np)^{-1/3}$, while the first term goes to $0$ because $H$ is continuous.