As mentioned in the comments, we need to fix isomorphisms $\newcommand{\D}{\mathscr{D}}\D_1 \xrightarrow{\sim} \D_1^{op}$ and $\D_2 \xrightarrow{\sim} \D_2^{op}$ for the question to make sense. In this case one can consider the gluings $\D = \D_1 \times_\varphi \D_2$ and $\D' = \D_2 \times_\psi \D_1$. Here, $\psi$ is the $\D_2$-$\D_1$-bimodule corresponding to $\varphi$ under the identification $\D_1^{op} \otimes \D_2 \xrightarrow{\sim} \D_2^{op} \otimes \D_1$.

In fact, $\D'$ is nothing but the opposite category of $\D$. One has canonical isomorphisms $$ \D' = \D_2 \times_\psi \D_1 \simeq \D_2^{op} \times_\varphi \D_1^{op} \simeq ( \D_1 \times_\varphi \D_2)^{op} = \D^{op} $$ where the last identification holds in general (without the duality assumptions), see Lemma 7.1 in Kuznetsov-Lunts.

Your construction seems to be fine, if by $\mu^{op}$ you mean the element $\mu' \in \psi^0(M_1, M_2)$ corresponding to $\mu \in \varphi^0(M_2, M_1)$ under the isomorphisms induced by duality. Further this functor $\D \to \D'$ is easily seen to be an equivalence, just by construction.

As mentioned in the comments, we need to fix isomorphisms $\newcommand{\D}{\mathscr{D}}\D_1 \xrightarrow{\sim} \D_1^{op}$ and $\D_2 \xrightarrow{\sim} \D_2^{op}$ for the question to make sense. They key point is that in this case, there is a canonical isomorphism $\D_1^{op} \otimes \D_2 \xrightarrow{\sim} \D_2^{op} \otimes \D_1$, inducing equivalences between the categories of $\D_1$-$\D_2$-bimodules and $\D_2$-$\D_1$-bimodules.

Hence one can consider the gluings $\D = \D_1 \times_\varphi \D_2$ and $\D' = \D_2 \times_\psi \D_1$, where $\psi$ is the $\D_2$-$\D_1$-bimodule corresponding to $\varphi$.

Indeed one has a functor $\D \to \D'$ as you described, where $\mu^{op}$ means the element of $\psi^0(M_1, M_2)$ corresponding to $\mu \in \varphi^0(M_2, M_1)$ under the isomorphisms induced by duality. Further this functor $\D \to \D'$ is clearly an equivalence, swapping the first two components and acting by duality on the third.

As mentioned in the comments, we need to fix isomorphisms $\newcommand{\D}{\mathscr{D}}\D_1 \xrightarrow{\sim} \D_1^{op}$ and $\D_2 \xrightarrow{\sim} \D_2^{op}$ for the question to make sense. In this case one can consider the gluings $\D = \D_1 \times_\varphi \D_2$ and $\D' = \D_2 \times_\psi \D_1$. Here, $\psi$ is the $\D_2$-$\D_1$-bimodule corresponding to $\varphi$ under the identification $\D_1^{op} \otimes \D_2 \xrightarrow{\sim} \D_2^{op} \otimes \D_1$.

In fact, $\D'$ is nothing but the opposite category of $\D$. One has canonical isomorphisms $$ \D' = \D_2 \times_\psi \D_1 \simeq \D_2^{op} \times_\varphi \D_1^{op} \simeq ( \D_1 \times_\varphi \D_2)^{op} = \D^{op} $$ where the last identification holds in general (without the duality assumptions), see Lemma 7.1 in Kuznetsov-Lunts.

As mentioned in the comments, we need to fix isomorphisms $\newcommand{\D}{\mathscr{D}}\D_1 \xrightarrow{\sim} \D_1^{op}$ and $\D_2 \xrightarrow{\sim} \D_2^{op}$ for the question to make sense. In this case one can consider the gluings $\D = \D_1 \times_\varphi \D_2$ and $\D' = \D_2 \times_\psi \D_1$. Here, $\psi$ is the $\D_2$-$\D_1$-bimodule corresponding to $\varphi$ under the identification $\D_1^{op} \otimes \D_2 \xrightarrow{\sim} \D_2^{op} \otimes \D_1$.

In fact, $\D'$ is nothing but the opposite category of $\D$. One has canonical isomorphisms $$ \D' = \D_2 \times_\psi \D_1 \simeq \D_2^{op} \times_\varphi \D_1^{op} \simeq ( \D_1 \times_\varphi \D_2)^{op} = \D^{op} $$ where the last identification holds in general (without the duality assumptions), see Lemma 7.1 in Kuznetsov-Lunts.

Your construction seems to be fine, if by $\mu^{op}$ you mean the element $\mu' \in \psi^0(M_1, M_2)$ corresponding to $\mu \in \varphi^0(M_2, M_1)$ under the isomorphisms induced by duality. Further this functor $\D \to \D'$ is easily seen to be an equivalence, just by construction.