Conventionally, we approximate a binomially-distributed variable $X\sim B(n, p)$ with the Poisson-distributed variable $Y\sim Po(np)$, with the mean of $X$ and $Y$ being identical.
However, we could also approximate $X$ with $Z\sim Po(np(1-p))$. While the mean of $X$ and $Z$ are different, the variances are identical.
That leads me to wonder if the "best" (perhaps under something like mean squared error) to approximate $X$ (with $n\gg 1$ and $p\ll \frac{1}{2}$) would be taking a Poisson distribution with $\lambda$ somewhere between $np(1-p)$ and $np$. So what would that optimal value $\lambda$ be?
I very much doubt that there's a closed-form solution for the optimal $\lambda$.
I tried an example: $n = 10$, $p = 1/10$. The objective is to minimize $$ g(\lambda) = \sum_{k=0}^\infty (\mathbb P(X = x) - \mathbb P(Y = x))^2$$ The optimal solution turns out to be the root of $$ 2 \,{\mathrm e}^{-2 \lambda} I_{1}\! \left(2 \lambda \right)-2 I_{0}\! \left(2 \lambda \right) {\mathrm e}^{-2 \lambda}+\frac{{\mathrm e}^{-\lambda} \left(\lambda^{10}+890 \lambda^{9}+319950 \lambda^{8}+60361200 \lambda^{7}+6503263200 \lambda^{6}+408316749120 \lambda^{5}+14624406014400 \lambda^{4}+279631499616000 \lambda^{3}+2473292401776000 \lambda^{2}+7029357352416000 \lambda -1405871470483200\right)}{18144000000000000} $$
which is approximately $1.042309798$, according to Maple. Note that this is greater than $np = 1$, while your $np(1-p) < 1$.