Twisted derived Morita theory of schemes It has been proved by Toën and Lunts-Schnürer that the dg category $\mathrm{L}_{qcoh}(X\times Y)$ of quasi-coherent sheaves over the product of two quasi-compact, quasi separated (and flat over a commutative ring $k$) schemes is quasi-equivalent to the dg-category of continuous dg functors $\mathbb{R}\underline{\mathrm{Hom}}_c(\mathrm{L}_{qcoh}(X),\mathrm{L}_{qcoh}(Y))$, thus providing a kernel for every triangulated functor between the derived categories of $X$ and $Y$ which is eligible for being of Fourier-Mukai type (i.e. commuting with direct sums). This equivalence also descends to the perfect dg subcategories and the internal (derived) Hom for dg categories above is defined by Toën in his paper https://arxiv.org/abs/math/0408337 by means of some dg bi-modules over the two dg categories.
What is known about whether this equivalence still holds when we consider the dg categories of complexes of (perfect or not) twisted sheaves? The paper https://arxiv.org/abs/1002.2599 by Toën proves that that the twisted categories have a generator, is that sufficient to claim some result à la Bondal-Van den Bergh saying that the box product of the two generators is the generator of the dg category of the product?
To summarize in one line: is it plausible to have a result like
$$
\mathbb{R}\underline{\mathrm{Hom}}_c(\mathrm{L}_{qcoh}(X,\alpha),\mathrm{L}_{qcoh}(Y,\beta)) \simeq \mathrm{L}_{qcoh}(X\times Y, \alpha^{-1}\boxtimes \beta) 
$$
eventually for perfect complexes? and if it is so plausible that someone already wrote it, where could I read it?
Thank you!
 A: I believe the answer is yes. Let attempt a sketch and give the appropriate references. The general strategy is to prove the main theorems of Ben-Zvi, Francis, Nadler in a twisted fashion. The main content of that paper is that the so-called "perfect stacks" are those algebro-geometric gadgets whose $L_{qcoh}$'s are exactly compactly generated and the dualizables and compacts coincide. As a result, we can keep track of these compact generators and prove that $L_{perf}(-)^c$ takes fiber products to tensor products (of dg-categories or what not). The desired result then follows from the fact that $L_{perf}(X)$ for $X$ again being a "perfect" stack (so that $L_{perf}(X) \otimes_{L_{perf}(k)} L_{perf}(Y) \simeq Fun_{L_{perf}(k)}( L_{perf}(X),  L_{perf}(Y))$. Here goes the sketch:


*

*In general we can twist by $\alpha: X \rightarrow Pr$, a linear category with descent [Antieau-Gepner, 6.2], but we restrict to the case where factors through $\alpha: X \rightarrow Alg$. For concreteness, this corresponds classically to twisting by an Azumaya algebra. The main point is: [Antieau-Gepner, 6.8], we want $L_{perf}^{\alpha}(U)$ for $U$ affine to be generated by a compact/dualizable/perfect object.

*Now, as you set it up, suppose $X, Y$ are qcqs schemes flat over a base $S$ (not sure if we need flat), let $\alpha, \beta$ be the Azumayas you want to twist by. In this case, we see that $L_{perf}^{\alpha/\beta}(-)$ for appropriate $(-)= X, Y, S$ are indeed compactly generated. We use the fact that this is true for affines and then use [Antieau-Gepner, 6.11] to see that we can glue the compact generators together in the twisted setting

*Next, we employ the argument of Ben-Zvi, Francis, Nadler, 4.7, but do it in a twisted way. One sees that all one needs is Ben-Zvi, Francis, Nadler, 4.6 where, in the twisted setting, one needs to prove that if $L_{perf}^{\alpha}(X) \otimes L_{perf}^{\beta}(Y) \simeq L_{perf}^{\alpha \otimes \beta}(X \times Y)$. Doing that only requires Ben-Zvi, Francis, Nadler, 3.24 to know that external products take compact generators to compact generators and that $L_{perf}^{\alpha \otimes \beta}(X \times Y)$ is generated by external products. As you suspect this is an argument ala Bondal-Van Den Bergh and you need only use the compact generation statements above

*Lastly, you need  Ben-Zvi, Francis, Nadler, 4.8 for self-duality, actually it seems that the dual of $L_{perf}^{\alpha}(X)$ is $L_{perf}^{\alpha^{-1}}(X)$, which explains that inversion in your formula. Anyway, the twisted categories are dualizable since they have a compact generator, and performing the diagram chase in that reference gives you the appropriate duality statement.
Admittedly, I haven't kept track of the twisting, and I might do so if I have more time to edit this answer
