Here are some suggestions, until somebody more knowledgeable appears.

1. Why does this definition make sense:
Analytic continuation is unique, and hence if there is analytic continuation to neighbourhood of $0$, and the formula is valid in finite dimension, then why not take it. Unfortunately, we have in general that
$$ \det (AB) \neq \det(A) \det(B).$$

2. Why does it possibly generalize the theory of the Riemann zeta function: 
If you take the circle with the Laplace operator $-\frac{\partial^2}{\partial^2 x}$ on $L^2[0,1]$ has eigenfunctions $\exp(2 \pi \textup{i} nx)$ for $n \in \mathbb{Z}$, you actually will get the Riemann zeta function as the spectral zeta function $\zeta_A(s) =  2 \zeta(2s)$.

3. Why does it not:
Certain properties are not shared by the general spectral zeta function. There is in general no functional equation relating $\zeta_A(s)$ to $\zeta_A( D-s)$ for some $D$, and there is no Euler product.


But here is also a quote from page 20, of "Ten physical applications of spectral zeta functions" from Elizalde, pg.20

Actually, a universal definition for the determinant [...] is still missing.

Also Singer and Ray refer in http://www.sciencedirect.com/science/article/pii/0001870871900454
to Seeley http://www.ams.org/mathscinet/search/publications.html?pg1=IID&s1=157950 , who has proven the regularity at $0$. He has a nice formula there:

$$A^s=i/2π\int_{Γ} λ^s(A−λ)^{−1}dλ$$