Assume that $0\to A_{\bullet}\to^{f} B_{\bullet}\to^g C_{\bullet}\to0$ is a short exact sequence of complexes of $R$-modules where $R$ is a commutative ring. This induces a homology exact sequence
$$\cdots\to H_{n}(A_{\bullet})\xrightarrow{f_*} H_{n}(B_{\bullet})\xrightarrow{g_*}H_{n}(C_{\bullet})\xrightarrow{\partial} H_{n-1}(A_{\bullet})\to\cdots,~~~~~~~~~~~(1)$$
where $f_*$, $g_*$ are the induced morphisms on homology and $\partial$ is the connecting morphism. Assume further that for every $n$, we have isomorphisms of $R$-modules $\phi_n: H_{n}(B_{\bullet})\cong H_n(D_{\bullet})$ and $\psi_n: H_n(C_{\bullet})\cong H_n(E_{\bullet})$.
Is it correct that the long exact sequence $(1)$ induces a long exact sequence of $R$-modules:
$$\cdots\to H_{n}(A_{\bullet})\xrightarrow{\phi_n\circ f_*} H_{n}(D_{\bullet})\xrightarrow{\psi_n\circ g_*\circ\phi_n^{-1}}H_{n}(E_{\bullet})\xrightarrow{\partial\circ\psi_n^{-1}} H_{n-1}(A_{\bullet})\to\cdots?~~~~~~~(2)$$
I believe yes and here is a simple proof why:
Proof:
A short computation demonstrates that at $H_n(D_{\bullet})$, we have: $$\ker(\psi_n\circ g_*\circ\phi_n^{-1})=(\phi_n^{-1})^{-1}(\ker(\psi_n\circ g_*))=\phi_n(g_*^{-1}(\ker(\psi_n)))=\phi_n(g_*^{-1}(0))=\phi_n(\ker g_*),$$ $$Im(\phi_n\circ f_*)=\phi_n(Im(f_*)).$$
From the exactness of the sequence $(1)$, we have $Im (f_*)=\ker(g_*)$. Hence, $\phi_n(Im(f_*))=\phi_n(\ker g_*)$. Hence, sequence $(2)$ is exact at $H_n(D_{\bullet})$.
Similarly, at $H_n(E_{\bullet})$, we have
$$Im(\psi_n\circ g_*\circ\phi_n^{-1})=\psi_n(g_*(H_n(B_{\bullet})))=\psi_n(Im(g_*)),$$ $$\ker(\partial\circ\psi_n^{-1})=\psi_n(\ker(\partial))$$.
Again, the exactness of $(1)$ implies $\psi_n(Im(g_*))=\psi_n(\ker(\partial))$. Hence, sequence (2) is exact at $H_n(E_{\bullet})$.
Finally, at $H_{n-1}(A_{\bullet})$, we have:
$$Im(\partial\circ\psi_n^{-1})=\partial(H_n(C_{\bullet}))=Im(\partial),$$ $$\ker(\phi_{n-1}\circ f_*)=f_*^{-1}(\ker(\phi_{n-1}))=f_*^{-1}(0)=\ker(f_*).$$
Hence, the exactness of sequence $(1)$ at $H_{n-1}(A_{\bullet})$ implies the exactness of sequence $(2)$ at $H_{n-1}(A_{\bullet})$.
On the basis of the above verification, I conclude that sequence $(2)$ is exact. QED