= Solution
Write $O=\int d[\bar\psi,\psi]e^{-\sum\bar\psi\psi}|\psi\rangle\langle\psi|$. For bosons the measure is $\prod_n d\operatorname{Re}\psi_n\,d\operatorname{Im}\psi_n/\pi$ over $\mathbb C$ in each mode, with $\bar\psi_n=\psi_n^*$. For fermions it is an ordered <Berezin integral> over independent $\bar\psi_n,\psi_n$; choose $\int d\bar\psi\,d\psi\,\bar\psi\psi=-1$, so $\int d\bar\psi\,d\psi\,e^{-\bar\psi\psi}=1$.
For one bosonic mode, $a|\psi\rangle=\psi|\psi\rangle$ and $\langle\psi|a=\partial_{\bar\psi}\langle\psi|$. <Integration by parts> in the Gaussian measure gives $aO=Oa$; the conjugate argument gives $a^\dagger O=Oa^\dagger$. Boundary terms vanish because of the Gaussian weight.
For one fermionic mode, put $n=a^\dagger a$ and move Grassmann coefficients to the left. The weighted projector is
$$
W=e^{-\bar\psi\psi}|\psi\rangle\langle\psi|=(1-n)-\psi a^\dagger+\bar\psi a-\bar\psi\psi I.
$$
Its ordinary <commutators> are $[a,W]=-a+\psi I$ and $[a^\dagger,W]=a^\dagger-\bar\psi I$. Their <Berezin integrals> vanish, so again $[a,O]=[a^\dagger,O]=0$. Equivalently the sole surviving coefficient in $\int W$ is $-\bar\psi\psi I$, giving $I$ directly.
The modes factorize. In the irreducible <Fock space> representation, commuting with every creation and annihilation operator makes $O$ a scalar multiple of the identity. Its vacuum matrix element is the normalized Gaussian integral, equal to one. Thus the <coherent-state resolution of identity> is
$$
\boxed{\int d[\bar\psi,\psi]e^{-\sum_n\bar\psi_n\psi_n}|\psi\rangle\langle\psi|=I}.
$$
For infinitely many modes, this argument first uses a finite-mode regulator.
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