Not a complete answer. First, here is an alternate derivation of the result in the finite-dimensional case which might be more enlightening. If $A$ is positive self-adjoint, we can write $A = \exp(L)$ for some self-adjoint $L$. This lets us define
$$A^s = \exp(sL)$$

for all real $s$. The trace
$$\text{tr}(A^s) = \sum_{i=1}^n \lambda_i^s = \zeta_A(s)$$

is then the zeta function associated to $A$ (I am getting rid of all the minus signs). Now, for small $\epsilon$ we can write
$$A^{s+\epsilon} = A^s A^{\epsilon} = A^s (1 + \epsilon L + O(\epsilon^2))$$

so it follows that
$$\zeta_A(0)' = \text{tr}(L).$$

But **Jacobi's identity** $\det \exp M = \exp \text{tr } M$ gives
$$\det A = \det \exp L = \exp \text{tr } L = \exp \zeta_A(0)'$$

and we conclude. 

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So what conceptual significance can we attach to the above? Well, it seems to me like we should think of the map $s \mapsto A^s$ as a representation of the Lie group $\mathbb{R}$, so the zeta function is the character of the corresponding representation. The derivative of the zeta function at zero gives the trace of the infinitesimal generator of this representation, $L$, which generates the abelian Lie algebra $\mathbb{R}$. And this is connected to the determinant of $A$ by Jacobi's identity.

So I think most of what needs explanation is Jacobi's identity. I freely admit that I do not have a good conceptual explanation of Jacobi's identity (beyond the fact that it's obvious for diagonalizable matrices). In <a href="http://qchu.wordpress.com/2009/08/23/newtons-sums-necklace-congruences-and-zeta-functions/">these</a> <a href="http://qchu.wordpress.com/2009/11/04/newtons-sums-necklace-congruences-and-zeta-functions-ii/">two</a> blog posts I attempted to meander towards a combinatorial proof of Jacobi's identity in the form
$$\det (I - At)^{-1} = \exp \text{tr } \log (I - At)^{-1}$$

(where $A$ was the adjacency matrix of a graph) but didn't quite succeed. There is a combinatorial proof of Jacobi's identity in the literature <a href="http://www.sciencedirect.com/science/article/pii/S016750600870699X">due to Foata</a> but I haven't gone through it in detail.