This is the picture, I have in mind.
Certainly analytic continuation gives a unique, well-defined value at $0$, if available, and generalizes Jacobi's formula (I like the logarithmic derivative variant best): $$ \frac{ \det(A(s))'}{\det(A(s))} = \operatorname{tr}\left(A^{-1}(s) A(s)'\right)$$ from the finite dimensional situation.
Hint: Take $A(s) = \exp( A^s)$ above and imagine $A$ as diagonalized operator.
Here is another variant $$\det(\exp(tA)) = 1 + t\ Tr(A) + O(t^2)$$
from Geometric interpretation of trace,
If you take the circle with the Laplace operator, you actually will get the Riemann zeta function as the spectral zeta function $\zeta_A(s) = \zeta(2s)$.
But the crucial difference between the above spectral zeta functions and the number theoretic ones is that the general zeta function does not admit a functional equation, in fact will not analytic continue to a meromorphic function of all of $\mathbb{C}$. They do analytically continue around a neighborhood of zero, if your eigenvalues do not grow to fast.
They do not grow to fast for pseudo differential operators on compact manifolds by a generalization of Weyl's law.