For the sake of the subsequent exploration, remember that if $E_i$ is a $\mathbb R$-vector space with $d:=\dim E_1\in\mathbb N$, $A\in\mathcal L(E_1,E_2)$ and $k:=\dim\mathcal N(A)$, then any basis $(e_1,\ldots,e_k)$ of $\mathcal N(A)$ can be extended to a basis of $(E_1)$ and $$f_i:=Ae_i\;\;\;\text{for }i\in\{k+1,\ldots,d\}$$ is a basis of $\mathcal R(A)$.
Now let $E_i,F_i$ be $\mathbb R$-vector spaces with $d_i:=\dim E_i\in\mathbb N$, $A_i\in\mathcal L(E_i,F_i)$, $k_i:=\dim\mathcal N(A_i)$ and $I$ be a finite nonempty set. By the aforementioned result, we can extend a basis $\left(e^{(i)}_1,\ldots,e^{(i)}_{k_i}\right)$ of $\mathcal N(A_i)$ to a basis $\left(e^{(i)}_1,\ldots,e^{(i)}_{d_i}\right)$ of $E_i$ and $$f^{(i)}_\alpha:=A_ie^{(i)}_\alpha\;\;\;\text{for }\alpha\in\{k_i+1,\ldots,d_i\}$$ is a basis of $\mathcal R(A_i)$. We know that $$e_\alpha:=\bigotimes_{i\in I}e^{(i)}_{\alpha_i}\;\;\;\text{for }\alpha\in\Gamma:=\{\alpha\in\mathbb N^I:\alpha_i\le d_i\;\text{for all }i\in I\}$$ is a basis of $\bigotimes_{i\in I}$. For convenience, let $$\Gamma_1:=\{\alpha\in\Gamma\mid\exists i\in I:\alpha_i\le k_i\}$$ and $$\Gamma_2:=\{\alpha\in\Gamma\mid\forall i\in I:k_i<\alpha_i\le d_i\}.$$
I would like to show that the rank of $\bigotimes_{i\in I}A_i\in\mathcal L(\bigotimes_{i\in I}E_i,\bigotimes_{i\in I}F_i)$ is equal to $\prod_{i\in I}\operatorname{rank}A_i$. By the aforementioned result, it would be enough to show that $(e_\alpha)_{\alpha\in\Gamma_1}$ is a basis of $\mathcal N(\bigotimes_{i\in I}A_i)$, since then the same argumentation as needed for this elementary result (see, for example, https://ltcconline.net/greenl/courses/203/MatrixOnVectors/kernelRange.htm) could be used to conclude that $$f_\alpha:=\left(\bigotimes_{i\in I}A_i\right)\bigotimes_{i\in I}e^{(i)}_{\alpha_i}=\bigotimes_{i\in I}A_ie^{(i)}_{\alpha_i}\;\;\;\text{for }\alpha\in\Gamma_2$$ is a basis of $\mathcal R(\bigotimes_{i\in I})$. (But maybe it's easier to show this directly.)
Actually, it's quite clear that the claim is true, but I'm really struggling in the notational complexity. Is there an easy way to show the claim?
Have you done the case $|I| = 2$? The general case follows from here, since there are natural vector space isomorphisms $$E_1 \otimes \cdots \otimes E_r \cong (E_1 \otimes \cdots \otimes E_{r-1}) \otimes E_r$$ $$F_1 \otimes \cdots \otimes F_r \cong (F_1 \otimes \cdots \otimes F_{r-1}) \otimes F_r$$ with respect to which the linear map $A_1 \otimes \cdots \otimes A_r$ identifies with $(A_1 \otimes \cdots \otimes A_{r-1}) \otimes A_r$.