Is the $0\times 0$ matrix (zero-times-zero matrix) a well-defined concept?

Well, one could say even more about the $0$ x $0$ matrix: Yes, it operates on the zero vector space (which contains only one single element = $0$). So, it maps $0$ to $0$ since there is no other possible image. Hence, it must be the identity mapping on the zero space, and therefore, it is its own inverse.

Suppose $A$ is a linear operator on a finite-dimensional $K$-vector space $V$. Say $\dim_K V = n$. It induces a natural linear map $\wedge^n A$ on the exterior power $\wedge^n V$. The latter is a $1$-dimensional $K$-vector space, so that $\wedge^n A$ is multiplication by a scalar. This scalar is the determinant of $A$. (This is the definition of determinant you'll find in Bourbaki.)

When $n = 0$, we have a natural identification $\wedge^0 V \cong K$, and the induced map $\wedge^0 A$ is (by construction) the identity map. I.e. multiplication by $1$. Thus $\det A = 1$.