This is explained in [Rickard, Jeremy. Derived equivalences as derived functors. J. London Math. Soc. (2) 43 (1991)], section 4.

An object $X$ in $D^b (\operatorname{\mathsf{Mod}} (A \otimes B^{\mathrm{op}}))$ is called a *two-sided tilting complex* if $$X \otimes^{\mathbb L}_B - \colon D^b (\operatorname{\mathsf{Mod}} B) \rightarrow D^b (\operatorname{\mathsf{Mod}} A)$$ is an equivalence of triangulated categories.

(In the question $A=B$, but I'll ignore that.)

Let $X$ be a two-sided tilting complex. Then by adjointness, $$\mathbb R \operatorname{Hom}_A(X,-) \colon D^b (\operatorname{\mathsf{Mod}} A) \rightarrow D^b (\operatorname{\mathsf{Mod}} B)$$ is also an equivalence of triangulated categories.
There are functorial isomorphisms
\begin{align}
\operatorname{Hom}_{D^b(\operatorname{\mathsf{Mod}} (A \otimes A^{\mathrm{op}}))}&(X \otimes^{\mathbb L}_B \mathbb R \operatorname{Hom}_A(X,A),-)\\
&\simeq \operatorname{Hom}_{D^b(\operatorname{\mathsf{Mod}} (B \otimes A^{\mathrm{op}}))}(\mathbb R \operatorname{Hom}_A(X,A),\mathbb R \operatorname{Hom}_A(X,-))\\
&\simeq \operatorname{Hom}_{D^b(\operatorname{\mathsf{Mod}} (A \otimes A^{\mathrm{op}}))}(A,-),
\end{align}
so $$X \otimes^{\mathbb L}_B \mathbb R \operatorname{Hom}_A(X,A) \\
\simeq A$$ in $D^b(\operatorname{\mathsf{Mod}} (A \otimes A^{\mathrm{op}}))$ by Yoneda's Lemma.

After first proving the isomorphism $$\mathbb R \operatorname{Hom}_A(X,A) \otimes^{\mathbb L}_A X \simeq \mathbb R \operatorname{Hom}_A(X,X \otimes^{\mathbb L}_B B),$$
a similar adjunction gives $$B \simeq \mathbb R \operatorname{Hom}_A(X,A) \otimes^{\mathbb L}_A X$$ in $D^b(\operatorname{\mathsf{Mod}} (B \otimes B^{\mathrm{op}}))$.