By Lemma \ref{lemma-functoriality-bc} we may replace $N$ and $N'$ by quasi-isomorphic bimodules. Thus we may assume $N$, resp.\ $N'$ has property (P) as differential graded $(A, B)$-bimodule, resp.\ $(B, C)$-bimodule, see Lemma \ref{lemma-bimodule-resolve}. We claim the lemma holds with the $(A, C)$-bimodule $N'' = N \otimes_B N'$. To prove this, it suffices to show that $$ N_B \otimes_B^\mathbf{L} N' \longrightarrow (N \otimes_B N')_C $$ is an isomorphism in $D(C, \text{d})$, see Lemma \ref{lemma-compose-tensor-functors-general}. \medskip\noindent Let $F_\bullet$ be the filtration on $N$ as in property (P) for bimodules. By Lemma \ref{lemma-bimodule-property-P-sequence} there is a short exact sequence $$ 0 \to \bigoplus\nolimits F_iN \to \bigoplus\nolimits F_iN \to N \to 0 $$ of differential graded $(A, B)$-bimodules which is split as a sequence of graded $(A, B)$-bimodules. A fortiori this is an admissible short exact sequence of differential graded $B$-modules and this produces a distinguished triangle $$ \bigoplus\nolimits F_iN_B \to \bigoplus\nolimits F_iN_B \to N_B \to \bigoplus\nolimits F_iN_B[1] $$ in $D(B, \text{d})$. Using that $- \otimes_B^\mathbf{L} N'$ is an exact functor of triangulated categories and commutes with direct sums and using that $- \otimes_B N'$ transforms admissible exact sequences into admissible exact sequences and commutes with direct sums we reduce to proving that $$ (F_pN)_B \otimes_B^\mathbf{L} N' \longrightarrow (F_pN)_B \otimes_B N' $$ is a quasi-isomorphism for all $p$. Repeating the argument with the short exact sequences of $(A, B)$-bimodules $$ 0 \to F_pN \to F_{p + 1}N \to F_{p + 1}N/F_pN \to 0 $$ which are split as graded $(A, B)$-bimodules we reduce to showing the same statement for $F_{p + 1}N/F_pN$. Since these modules are direct sums of shifts of $(A \otimes_R B)_B$ we reduce to showing that $$ (A \otimes_R B)_B \otimes_B^\mathbf{L} N' \longrightarrow (A \otimes_R B)_B \otimes_B N' $$ is a quasi-isomorphism. \medskip\noindent Choose a filtration $F_\bullet$ on $N'$ as in property (P) for bimodules. Choose a quasi-isomorphism $P \to (A \otimes_R B)_B$ of differential graded $B$-modules where $P$ has property (P). We have to show that $P \otimes_B N' \to (A \otimes_R B)_B \otimes_B N'$ is a quasi-isomorphism because $P \otimes_B N'$ represents $(A \otimes_R B)_B \otimes_B^\mathbf{L} N'$ in $D(C, \text{d})$ by the construction in Lemma \ref{lemma-derived-bc}. As $N' = \colim F_pN'$ we find that it suffices to show that $P \otimes_B F_pN' \to (A \otimes_R B)_B \otimes_B F_pN'$ is a quasi-isomorphism. Using the short exact sequences $0 \to F_pN' \to F_{p + 1}N' \to F_{p + 1}N'/F_pN' \to 0$ which are split as graded $(B, C)$-bimodules we reduce to showing $P \otimes_B F_{p + 1}N'/F_pN' \to (A \otimes_R B)_B \otimes_B F_{p + 1}N'/F_pN'$ is a quasi-isomorphism for all $p$. Then finally using that $F_{p + 1}N'/F_pN'$ is a direct sum of shifts of ${}_B(B \otimes_R C)_C$ we conclude that it suffices to show that $$ P \otimes_B {}_B(B \otimes_R C)_C \to (A \otimes_R B)_B \otimes_B {}_B(B \otimes_R C)_C $$ is a quasi-isomorphism. Since $P \to (A \otimes_R B)_B$ is a resolution by a module satisfying property (P) this map of differential graded $C$-modules represents the morphism (\ref{equation-plain-versus-derived-algebras}) in $D(C, \text{d})$ and the proof is complete.