This question arise from the comparision of the reconstruction theorems of Bondal-Orlov and Balmer and is inspired by Shizhuo Zhang's mathoverflow question: How to unify various reconstruction theorems (Gabriel-Rosenberg, Tannaka,Balmers)

On the one hand, A. Bondal and D. Orlov in http://arxiv.org/abs/alg-geom/9712029 proved their celebrated reconstruction theorem: Let $X$ be a smooth projective variety such that the canonical bundle $\omega_X$ is either ample or anti-ample, and let $Y$ be any projective variety. If $D^b_{\text{coh}}(X)\cong D^b_{\text{coh}}(Y)$ as triangulated categories, then $X\cong Y$.

In fact, they reconstruct $X$ from $D^b_{\text{coh}}(X)$ and their construction heavily use the $\textit{Serre functor}$ S. In general, for a $k$-linear category $\mathcal{C}$, an equivalence $S: \mathcal{C}\rightarrow \mathcal{C}$ is called a Serre functor for $\mathcal{C}$ if their is a natural, bifunctorial isomorphisms $$ \varphi_{A,B}: \text{Hom}_{\mathcal{C}}(A,B)\xrightarrow{\sim} \text{Hom}_{\mathcal{C}}(B,SA)^{\vee} $$ for every $A,B$. Here $(\bullet)^\vee$ denotes the $k$-dual. When $\mathcal{C}=D^b_{\text{coh}}(X)$, the Serre functor $S_X$ is given by $$ S_X\mathcal{E}=\mathcal{E}\otimes \omega_X[\dim X] $$ i.e tensor product with the canonical bundle and shift by $\dim X$ in $D^b_{\text{coh}}(X)$. We can refer to Section 4 of A. Caldararu's lecture notes http://arxiv.org/abs/math/0501094 for an excellent introduction. We notice that Bondal-Orlov reconstruction does not involve the tensor structure of $D^b_{\text{coh}}(X)$ besides the definition of the Serre functor.

(As pointed out by Piotr and Qiaochu in the comments, the Serre functor is unique, hence is part of the data of $D^b_{\text{coh}}(X)$ and is not an extra data.)

On the other hand, P. Balmer in http://arxiv.org/abs/math/0111049 proved another reconstruction theorem: Let $X$ be a noetherian scheme, then we can reconstruct $X$ from the $\textit{tensor triangulated category}$ $(D^{\text{perf}}(X),\otimes^L_{\mathcal{O}_X})$. This reconstruction uses the extra structure of $D^{\text{perf}}(X)$: the tensor product, and it applies for more general $X$ than Bondal-Orlov.

Now we can look at the case when the two theorems overlap: If $X$ is smooth projective with $\omega_X$ ample or anti-ample, then $D^{\text{perf}}(X)\cong D^b_{\text{coh}}(X)$. Now from $D^b_{\text{coh}}(X)$ either using the Serre functor or using the tensor structure. It seems that there are some redundancy here. In fact we have $$ D^b_{\text{coh}}(X) \xrightarrow [\text{reconstruction}]{\text{Bondal-Orlov}}X\rightarrow (D^b_{\text{coh}}(X),\otimes^L_{\mathcal{O}_X})\xrightarrow [\text{reconstruction}]{\text{Balmer}} X\rightarrow D^b_{\text{coh}}(X) \ldots $$ Hence one may expect a direct construction of the tensor structure $(D^b_{\text{coh}}(X),\otimes^L_{\mathcal{O}_X})$ just from $D^b_{\text{coh}}(X)$ itself, considered as a triangulated category.

$\textbf{My question}$ is: If $X$ is smooth projective variety with $\omega_X$ ample or anti-ample, could we define the tensor product structure on $D^b_{\text{coh}}(X)$ just from the triangulated cateogry structure on $D^b_{\text{coh}}(X)$?

In the other direction we have $\textbf{a related question}$: What is the role of the Serre functor $S_X$ and the canonical bundle $\omega_X$ in the the tensor triangulated category $(D^b_{\text{coh}}(X),\otimes^L_{\mathcal{O}_X})$?