If $R$ is a ring, $M$ is a right $R$-module, and $N$ is a left $R$-module, the derived tensor product $M \otimes_R^{\mathbf{L}} N$ is computed by choosing projective resolutions $P_* \to M$ and $Q_* \to N$ and computing any of $M \otimes_R Q$, $P \otimes_R Q$, or $P \otimes_R N$.
I'm trying to understand the derived tensor product of bimodules. If $M$ is an $R$-$S$ bimodule and $N$ is an $S$-$T$ bimodule, then I can compute the derived tensor product $M \otimes_S^{\mathbf{L}} N$ by choosing projective resolutions $P_* \to M$ and $Q_* \to N$ and computing $P \otimes_S Q$. My question is: is it sufficient to resolve just one of $M$ and $N$?
The approach I want to take is to show that tensoring with a projective bimodule $P$ (or, more generally, a levelwise projective complex of bimodules) preserves quasi-isomorphisms. I can almost argue this: if $f$ is a quasi-isomorphism, then $\mathrm{cone}(f)$ is exact, so $\mathrm{cone}(P \otimes f) \cong P \otimes \mathrm{cone}(f)$ is exact since $P$ is (levelwise) projective, and thus $P \otimes f$ is a quasi-isomorphism. If I'm not mistaken, this works if everything is just a module over the same ring, but for bimodules there's an issue: $P$ being projective as a bimodule does not (to my knowledge) imply that $P$ is flat as a right module, and so I can't conclude that $P \otimes \mathrm{cone}(f)$ is exact.
I've looked over the proof of Theorem 2.7.2 in Weibel (balancing $\mathrm{Tor})$, and I believe it's very similar to my approach and has the same issue: it uses that tensoring with a projective module is exact. Am I wrong to be resolving $M$ and $N$ by projective bimodules instead of modules that are just projective on one side? My understanding is that derived tensor should be a functor out of the product of the derived categories of $R$-$S$ and $S$-$T$ bimodules; I compute derived functors by cofibrantly replacing the objects, and in the projective model structure on chain complexes, the cofibrant objects are the complexes which are levelwise projective (as bimodules).