What can be said is that if $C,C'$ have degree-wise finite filtrations and if $f: C \to C'$ is a filtration preserving chain map such that there is $r$ such that $E^r(f): E^r_{ij}(C) \to E^r_{ij}(C')$ is an isomorphism for all $i,j$ then $f$ is a quasi-isomorphm. This can be found in Brown: Cohomology of Groups, VII, Prop. 2.6.
Special cases where $E^\infty(C) \cong E^\infty(C')$ implies $H_\ast(C) \cong H_\ast(C')$ include:
The spectral sequence degenerates, e.g. $E^\infty_{ij}=0$ if $j \neq 0$ because in this case $H^i(C)=E^\infty_{i,0}$.
If $C,C'$ are complexes of modules over a ring $R$ and $E^\infty(C)$ consists of projective $R$-modules. In this case there is no extension problem and we have $H^n(C)=\oplus_{i+j=n}E^\infty_{ij}(C)$.
Here it is essential to consider the $E^\infty$-term since it can happen that $E^r(C) \cong E^r(C')$ but $E^{r+1}(C) \not\cong E^{r+1}(C')$. As an example take the LHS spectral sequence of the group extension
$$0 \to \mathbb{Z}/2 \to G \to \mathbb{Z}/2 \oplus \mathbb{Z}/2 \to 0.$$
Both, the dihedral group $D_8$ and the quaternion group $Q_8$ fit into this extension. Hence the spectral sequences have the same $E_2$-term but their $E_3$-terms differ.
