This is a fantastic question, as it gets at the heart of a common misconception. The answer to the question "Why does $np$ have to be of moderate size?" is that it does not! The only thing that is needed for Poisson approximation is that $p$ is small.
To make this precise, we need a way to measure the quality of a probabilistic approximation. If $X$ and $Y$ are random variables, we define the total variation distance between them to be
$$
d_{\text{TV}}(X, Y)=\sup_{A} |P(X\in A)-P(Y\in A)|
$$
If this distance is small, say it is at most $\epsilon$, then you can use the distribution of $Y$ to calculation the probability of an event involving $X$, and have an absolute error of at most $\epsilon$.
Theorem Let $X\sim \text{Binomial}(n,p)$, let $\lambda=np$, and let $Y\sim \text{Poisson}(\lambda)$. Then
$$
d_\text{TV}(X,Y)\le \min\{p,np^2\}\tag1
$$
Therefore, as long as $p$ is small, the quality of the Poisson approximation is good.
This same idea is true in greater generality. For a Binomial distribution, there are $n$ independent and equiprobable events; we can drop the requirement of equiprobability. Let $E_1,\dots,E_n$ be independent events, with $P(E_i)=p_i$ for $i\in \{1,\dots,n\}$. If we let $X$ be the number of events which occur, and let $Z$ be a Poisson random variable with the same mean as $X$, meaning that $Z\sim \text{Poisson}(\lambda)$ with $\lambda=\sum_{i=1}^n p_i$, then
$$
d_\text{TV}(X, Z)\le \min\left\{\frac1{\lambda}\sum_{i=1}^n p_i^2, \,\,\sum_{i=1}^n p_i^2\right\}\tag2
$$
In the case where $p_1=\dots=p_n=p$, we recover $(1)$. For a proof of $(2)$, see Approximate Computations of Expectations, by Charles Stein. Equation $(2)$ appears on page 89.
Some comments on the semantic parts of your questions.
"Moderate" is not a precise term; it just means "neither larger nor small," but large and small are subjective.
When people say that $\lambda=np$ is constant as $n\to\infty$, it just means that $n$ is getting larger, and for each particular instance of $n$, the corresponding value of $p$ is equal to $\lambda/n$. Saying
The Binomial$(n,p)$ distribution as $n\to\infty$ with $\lambda=np$ constant.
means the exact same thing as
The Binomial$(n,\tfrac{\lambda}n)$ distribution as $n\to\infty$.
So, Binomial$(10,0.5)$ is pretty close to a Poisson$(5)$ distribution, Binomial$(50,0.1)$ is even closer, Binomial$(500,0.01)$ is extremely close, etc.
