Let $GL(n) := \mathrm{GL}(n, \mathbb{C})$ - be space of invertible $n\times n$ matrices over $\mathbb{C}$, i.e. matrix Lie group.
Let $H := GL(n) \otimes GL(n)$ be a group with multiplication given by $(a \otimes b)\cdot (c \otimes d) = ac \otimes bd$. We embed $H$ into $G:=GL(n^2)$ as a subgroup via Kronecker product of matrices: $$(A \otimes B)_{ij} = A_{i/n,i\% n} B_{j/n,j\%n}$$ for each $i,j \in [0,n^2-1]$ where $\%$ is taking modulo and $/$ is an integer division. Note that $\det(A\otimes B) = (\det(A)\det(B))^n$ guarantees the correctness of an embedding.
Clearly, $H$ is closed under multiplication of $G$, i.e. $H$ is a subgroup of $G$.
What I am trying to understand is if $H$ is closed in $G$ in topological sense. I need this in order to state that $H$ is a closed subgroup of a Lie group $G$.
The question was essentially answered by user The Phoenix in comments. I am posting detailed answer for non-experienced reader (such as me).
Denote $M_n$ as the set of all $n \times n$ matrices over $\mathbb{C}$. To prove that $H$ is closed in $G$ as topological set we recall definition:
Here $T = M_{n^2}$, $T' = GL(n^2)$ and $V = H$. Since $H = (M_n \otimes M_n) \cap T'$ (due to properties of determinants mentioned in question) it is enough to show that $M_n \otimes M_n$ is closed in $M_{n^2}$. Indeed, topologically $M_n \cong \mathbb{C}^{n^2}$ and the subset $M_n \otimes M_n \subset M_{n^2} \cong \mathbb{C}^{n^4}$, so called Segre embedding, defined by system of finitely many polynomial equations (certain quadratic equations) $P_1 = 0,\ldots,P_\ell = 0$. Polynomial is a continuous function, then each set $P_i^{-1}(\{0\})$ is closed and intersection of these closed sets $\bigcap_i P_i^{-1}(\{0\}) = M_n \otimes M_n$ is also closed.