Proof that $\text{det}(AB) = \text{det}(A)\text{det}(B)$ without explicit expression for $\text{det}$

Disclaimer: I am cheating a little bit here because

1. In some sense, I just proved that two different definitions of the determinant are equal (namely the one that the OP gave and the one using multilinear forms).

2. One would still have to prove that the determinant as defined by the OP is continuous (which must be true). Perhaps one can use that the characteristic polynomial varies continuously as you move continuously through $\text{End}(V)$.

Let $\Lambda^n(V)$ denote the space of alternating $n$-multilinear transformations. For each $T\in\text{End}(V)$ there is a pull-back map $T^*:\Lambda^n(V)\rightarrow\Lambda^n(V)$ defined by $$ T^*(\omega)(v_1,\ldots,v_n)=\omega(T(v_1),\ldots,T(v_n)) ,$$ for each $\omega\in\Lambda^n(V)$ and $v_1,\ldots,v_n\in V$. Since $\Lambda^n(V)$ is $1$-dimensional, one can define $$\widetilde{\det}:\text{End}(V)\rightarrow \mathbb{C}$$ by setting $$ T^*(\omega)=\widetilde{\det}(T)\omega,$$ for each $\omega\in\Lambda^n(V)$. Then, by @hm2020's answer, it is clear that $\widetilde{\det}(AB)=\widetilde{\det}(A)\widetilde{\det}(B)$ (it follows from the definition at once). Now, for each diagonalizable $A\in\text{End}(V)$, take a basis of eigenvectors $\{e_i\}_{i=1}^n$ with eigenvalues $\{\lambda_i\}_{i=1}^n$ and $n$ vectors $v_k=\displaystyle\sum_{i=1}^n v^i_ke_i\in V$, $k\in\{1,\ldots,n\}$. It follows that \begin{equation*}\begin{split}T^*(\omega)(v_1,\ldots,v_n)&=\omega(T(v_1),\ldots,T(v_n))\\&=\omega(\sum v^i_1\lambda_ie_i,\ldots,\sum v^i_n\lambda_ie_i)\\ &=\left(\displaystyle\prod_{i=1}^n\lambda_i\right)(\omega(v_1,\ldots,v_n))\\ &=\det(T)\omega(v_1,\ldots,v_n), \end{split}\end{equation*} where we used that $\omega$ is alternating. It follows that $\det=\widetilde{\det}$ on the set of diagonalizable matrices, which is dense in $\text{End}(V)$. Since both of them agree on a dense set and are continuous (and the codomain is Hausdorff), we have $\det(T)=\widetilde{\det}(T)$ for every $T\in\text{End}(V)$. In particular, for every $A,B\in\text{End}(V)$, there holds $$ \det(AB)=\widetilde{\det}(AB)=\widetilde{\det}(A)\widetilde{\det}(B)=\det(A)\det(B).$$

Question: Proof that $det(AB)=det(A)det(B)$ without explicit expression for $det$.

Answer: A "Bourbaki-proof": You may use the exterior product to prove the formula for any two matrices $A,B$ (of the same rank) over any commutative unital ring:

Example: Let $k$ be any commutative unital ring. If $\phi$ is the matrix

\begin{align*} \phi= \begin{pmatrix} a & b \\ c & d \end{pmatrix} \end{align*}

and you view $\phi:V:=k\{e_1,e_2\} \rightarrow k\{e_1,e_2\}$ as a $k$-linear endomorphism, it follows by the definition of the exterior product that you get an endomorphism

$$\wedge^2(\phi): \wedge^2 V \rightarrow \wedge^2 V$$

defined by

$$\wedge^2 \phi(e_1\wedge e_2)=\phi(e_1)\wedge \phi(e_2)= (ae_1+ce_2)\wedge (be_1+de_2)=(ad-bd)e_1 \wedge e_2:= $$

$$det(\phi)e_1 \wedge e_2.$$

Here we have used the property that $e_2 \wedge e_1=-e_1 \wedge e_2$ in the exterior product $\wedge^2 V$. More generally if $\phi$ is multiplication with $A:=(a_{ij})$ - an $n\times n$-matrix with coefficients in $k$, and $V:=k\{e_1,..,e_n\}$ with $\phi: V \rightarrow V$ it follows we get the formula

$$\wedge^n \phi(e_1\wedge \cdots \wedge e_n):=\phi(e_1)\wedge \cdots \wedge \phi(e_n)=det(A)e_1\wedge \cdots \wedge e_n.$$

Here we define $det(A)$ as in your thread.

Principle: Taking the $n$'th exterior power of an $n \times n$-matrix $A$ is an "abstract" construction of the determinant $det(A)$.

Example: Let $E:=k\{e_1,..,e_n\}$ be a free $k$-module of rank $n$ and let $\def\End{\operatorname{End}} A,B\in \End_R(E)$ be $n \times n$-matrices with coefficients in $k$ ($k$ any commutative unital ring). It follows $$\wedge^n A, \wedge^n B \in \End_R(\wedge^n E)$$ are $k$-linear endomorphisms of the rank one free $k$-module $\wedge^n E \cong k e_1\wedge \cdots \wedge e_n:=k e$ with $e:=e_1\wedge \cdots \wedge e_n$. It has the property that

$$\wedge^n A(ue) = \det(A) ue$$

is "multiplication with the element $\det(A)\in k$".

From this it follows since $\wedge^n$ is a functor (see the below link) that

$$ \det(AB) := \wedge^n (A \circ B) = \wedge^n A \circ \wedge^n B := \det(A) \det(B). $$

With the formulation using the exterior product, it follows $\wedge^n(\phi)(u)=det(A)u$ is a "theorem". Here we define (as in your thread)

$$D1.\text{ }det(A) = \sum_{\sigma \in S_N}\prod_{i=1}^N a_{i, \sigma(i)}.$$

Hence if we define the determinant using the exterior product, it follows formula D1 must be proved. Conversely if we define the determinant using D1, it follows the multiplicativity of the determinant must be proved.

Prove that $\det(T) = \det(T_{w_{1}})\det(T_{w_{1}})\cdots \det(T_{w_{k}})$ where $W_{i}$ are $T$-invariant subspaces of $V$ - Trouble at last step.