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 http://mathoverflow.net/questions/13526/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.