= Solution
Interpret $L=-\Box$ as the minimally coupled scalar <positive Laplace-Beltrami operator> on the smooth <closed manifold>. Writing $dV$ for its metric volume measure (the <Riemannian volume form> on an oriented chart), integration by parts gives $\langle u,Lu\rangle=\int_M|\nabla u|^2dV\ge0$. The <compact elliptic spectral theorem> supplies an orthonormal basis $u_j$ with $Lu_j=\lambda_j u_j$, real nonnegative <eigenvalues> tending to infinity and finite multiplicities. Solve this eigenvalue problem to obtain the <spectral zeta function>
$$
\boxed{\zeta_L(s)=\sum_{\lambda_j>0}\lambda_j^{-s},\qquad \operatorname{Re}s>d/2.}
$$
The zero <eigenvalues> are excluded, not assigned inverse powers. Their multiplicity $n_0$ is the number of connected components: $Lu=0$ implies $\int|\nabla u|^2dV=0$, so $u$ is constant on each component.
An equivalent calculation uses the <spectral expansion of the Riemannian heat kernel>,
$$
K(t;x,y)=\sum_j e^{-t\lambda_j}u_j(x)\overline{u_j(y)},\qquad
K(t)=\int_M K(t;x,x)dV=\sum_j e^{-t\lambda_j}.
$$
Since $\int_0^\infty t^{s-1}e^{-t\lambda}\,dt=\Gamma(s)\lambda^{-s}$ for $\lambda>0$, the <Mellin transform> gives
$$
\boxed{\zeta_L(s)=\frac1{\Gamma(s)}\int_0^\infty t^{s-1}(K(t)-n_0)\,dt.}
$$
The reduced trace decays exponentially at large $t$. At small $t$, subtract a finite number of terms of the <heat kernel expansion>, integrate those terms explicitly, and then integrate the remainder. Increasing the number of subtractions extends the formula successively to further left half-planes, producing the <meromorphic continuation> of $\zeta_L$ rather than substituting $s=0$ in its divergent defining series.
For completeness, the needed curvature coefficient comes from the <heat-kernel transport equations>. In <normal coordinates> $\xi$ centered at $y$, the <Riemannian volume form> has density $J(\xi)=1-\tfrac16R_{ab}(y)\xi^a\xi^b+O(|\xi|^3)$. The leading Gaussian amplitude is therefore
$$
u_0(x,y)=J(\xi)^{-1/2}=1+\frac1{12}R_{ab}(y)\xi^a\xi^b+O(|\xi|^3).
$$
Substitution of the Gaussian amplitude series into the heat equation gives the first diagonal transport relation $u_1(y,y)=\Box_xu_0(x,y)|_{x=y}=R(y)/6$. The <curvature coefficient of the scalar heat kernel> is consequently
$$
K(t;x,x)\sim\frac1{(4\pi t)^{d/2}}\left(1+\frac t6R(x)+O(t^2)\right).
$$
In dimension two this implies
$$
K(t)=\frac{\operatorname{Area}(M)}{4\pi t}+\frac1{24\pi}\int_M R\,dV+O(t).
$$
Set $A_0=\operatorname{Area}(M)/(4\pi)$ and $A_1=\int_M R\,dV/(24\pi)$. Splitting the Mellin integral at $t=1$ and subtracting these two heat terms gives, near $s=0$,
$$
\zeta_L(s)=\frac1{\Gamma(s)}\left[\frac{A_0}{s-1}+\frac{A_1-n_0}{s}+H(s)\right],
$$
where $H$ is holomorphic for $\operatorname{Re}s>-1$. As $1/\Gamma(s)=s+O(s^2)$, only the second term survives at zero. Thus the <zero-mode correction to the spectral zeta value> is
$$
\boxed{\zeta_L(0)=\frac1{24\pi}\int_M R\,dV-n_0.}
$$
The <Gauss-Bonnet theorem>, with $R$ twice the Gaussian curvature, makes this $\boxed{\zeta_L(0)=\chi(M)/6-n_0}$. For a connected surface $n_0=1$, so the answer is $\chi(M)/6-1$; a sphere gives $-2/3$, and a flat torus gives $-1$.
The purely local heat coefficient is $A_1=\int R\,dV/(24\pi)$. Sometimes a formal formula for $\zeta(0)$ quotes this coefficient with zero modes left implicit. For the actual positive-spectrum <spectral zeta function> defined above, the compact scalar problem necessarily has the additional subtraction $-n_0$. A strictly positive shifted operator is a different problem and has its own heat coefficients.
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