Two different sources use two different sets of assumptions for the Poisson distribution as a limiting case of the binomial distribution. This video sets $$n\rightarrow\infty, p\rightarrow0, np\rightarrow\lambda$$
And this sets $$n\rightarrow\infty, np=\lambda$$
The derivation remains the same. Is there any difference between the implications of these assumptions, and if so, what is it? If there is no difference, can we derive one from the other?
On
There is the estimate of the total variation norm $$\|Bin(n, p) - Pois(np)\| \leq n\|Ber(p) - Pois(p)\| \leq np^2.$$ This holds for every $n \in \mathbb{N}, p \in [0, 1]$. See section 3.6 of https://services.math.duke.edu/~rtd/PTE/PTE5_011119.pdf. Now if you consider a sequence $p_n$ with $np_n \to \lambda$, then a natural condition for the above norm to go to $0$ is that $p_n \to 0$. But the inequalities behind the estimate let's you formulate a more general condition under which sums of Bernoullis converge to a Poisson.
There is no difference. Assuming $\lambda > 0$, the condition $np = \lambda$ implies that as $n \to \infty$, we must have $p \to 0^+$.
Conversely, we only require $np \to \lambda$ asymptotically, rather than identically, as long as both $n \to \infty$ and $p \to 0^+$. (Some sources omit the $+$ because it is understood that $0 < p < 1$.) In fact, this (first) characterization is slightly more general than the second, but in practice, nothing is lost by assuming $np = \lambda$.