You ask for a relation between the Green's function of the single-particle Hamiltonian and the Green's function of the many-particle Hamiltonian, in the case of non-interacting particles (fermions or bosons). Let me try to explain that the retarded Green's functions are identical.
• For the single-particle Hamiltonian $H(x)=-\partial_x^2+V(x)$ the retarded Green's function is defined by $$i\partial_t G(x,x';t,t')=H(x)G(x,x';t,t'),\;\;t>t',$$ with the condition that $G(x,x';t,t')\equiv 0$ for $t<t'$ and $$\lim_{t\downarrow t'}G(x,x';t,t')=-i\delta(x'-x).$$
• The retarded many-particle Green's function is defined as the ground state expectation value $\langle\cdots\rangle$ of the (anti-)commutator of field operators $\hat\psi(x,t)$, $${\cal G}(x,x';t,t')=-i\langle\hat\psi(x,t)\hat\psi^\dagger(x',t')\pm\hat\psi^\dagger(x',t')\hat\psi(x,t)\rangle\theta(t-t').$$ The function $\theta(t)$ is the unit step function, the $+$ sign is for fermions and the $-$ sign for bosons. The field operator satisfies the operator equation $$i\partial_t\hat\psi(x,t)=[\hat\psi(x,t),\hat{\cal H}],$$ with $\hat{\cal H}$ the many-particle Hamiltonian operator and $[\cdot,\cdot]$ the commutator. Note also the equal-time (anti-commutation) relation $$\hat\psi(x,t)\hat\psi^\dagger(x',t)\pm\hat\psi^\dagger(x',t)\hat\psi(x,t)=\delta(x-x').$$
• Now let us compare the two functions $G(x,x';t,t')$ and ${\cal G}(x,x';t,t')$. Both vanish for $t<t'$ and both satisfy the delta-function limit when $t\downarrow t'$. But in general the function ${\cal G}$ is not the Green's function of any differential equation, unlike $G$.
However, for non-interacting particles we have $$[\hat\psi(x,t),\hat{\cal H}]=(-\partial_x^2+V(x))\hat\psi(x,t)=H(x)\hat\psi(x,t),$$ hence ${\cal G}$ satisfies the same differential equation $$i\partial_t {\cal G}(x,x';t,t')=H(x){\cal G}(x,x';t,t'),$$ as $G$ and we conclude that they are the very same function.