There are probably too many such $(M,g,I_+,I_-)$ to really expect a 'classification'.

For instance, consider the case when a complex manifold $(M,I_+)$ has real dimension $4$, and the $I_+$-holomorphic tangent bundle $T'M$ splits as the sum of two holomorphic line subbundles $T'M = L_1\oplus L_2$. In this case, one can define an $I_-$ by reversing $I_+$ on the line bundle $L_2$. Now let $\Omega_i$ for $i=1,2$ be a smooth real (1,1)-form of complex rank $1$ whose kernel is $L_{3-i}$ and is such that $\Omega_i$ restricts to $L_i$ to be a positive $(1,1)$-form (with respect to $I_+$). Let $g$ be the Hermitian metric (with respect to $I_+$) that is associated to the positive $(1,1)$-form $\Omega_+ = \Omega_1 + \Omega_2$ (with respect to $I_+$). Then the $2$-form $\Omega_- = \Omega_1 - \Omega_2$ is a positive $(1,1)$-form with respect to $I_-$ and it belongs to the same associated metric $g$.

It's now easy to check that the two Lee forms satisfy $\theta_+=\theta_-$. Note that replacing $\Omega_1$ and $\Omega_2$ by $f_1\Omega_1$ and $f_2\Omega_2$ where the $f_i$ are any positive (smooth) functions on $M$ will give other examples with equal Lee forms.

It's not hard to show that, in dimension $4$ this Ansatz the only way that such examples can arise. In higher dimensions, I expect that the analysis is more complicated, but, again, there will be too many examples to really classify because having equal Lee forms is a very underdetermined set of PDE on the data $(M,g,I_+,I_-)$.

There are probably too many such $(M,g,I_+,I_-)$ to really expect a 'classification'.

For instance, consider the case when a complex manifold $(M,I_+)$ has real dimension $4$, and the $I_+$-holomorphic tangent bundle $T'M$ splits as the sum of two holomorphic line subbundles $T'M = L_1\oplus L_2$. In this case, one can define an $I_-$ by reversing $I_+$ on the line bundle $L_2$. Now let $\Omega_i$ for $i=1,2$ be a smooth real (1,1)-form of complex rank $1$ whose kernel is $L_{3-i}$ and is such that $\Omega_i$ restricts to $L_i$ to be a positive $(1,1)$-form (with respect to $I_+$). Let $g$ be the Hermitian metric (with respect to $I_+$) that is associated to the positive $(1,1)$-form $\Omega_+ = \Omega_1 + \Omega_2$ (with respect to $I_+$). Then the $2$-form $\Omega_- = \Omega_1 - \Omega_2$ is a positive $(1,1)$-form with respect to $I_-$ and it belongs to the same associated metric $g$.

It is easy to check, using local coordinates, that the two associated Lee forms are equal: Fix a point $p\in M$ and choose an $I_+$-holomorphic coordinate chart $(z^1,z^2):U\to\mathbb{C}^2$ on a $p$-neighborhood $U\subset M$ such that $\mathrm{d}z^i=0$ defines the subbundle $L_{3-i}$. Note that $\bigl(z^1,\overline{z^2}\bigr):U\to\mathbb{C}^2$ is an $I_-$-holomorphic chart on $U$. There exist functions $u_i$ on $U$ such that $$ \Omega_{\pm} = \tfrac{\imath}2\,\mathrm{e}^{u_1}\,\mathrm{d}z^1\wedge\mathrm{d}\overline{z^1}\pm \tfrac{\imath}2\,\mathrm{e}^{u_2}\,\mathrm{d}z^2\wedge\mathrm{d}\overline{z^2}, $$ and their common Hermitian metric is $g = \mathrm{e}^{u_1}\,\mathrm{d}z^1\circ\mathrm{d}\overline{z^1}+ \mathrm{e}^{u_2}\,\mathrm{d}z^2\circ\mathrm{d}\overline{z^2}$. Then $\mathrm{d}\Omega_\pm = \theta\wedge\Omega_\pm$ where $$ \theta = \frac{\partial u_2}{\partial z^1}\,\mathrm{d}z^1 +\frac{\partial u_2}{\partial \overline{z^1}}\,\mathrm{d}\overline{z^1} +\frac{\partial u_1}{\partial z^2}\,\mathrm{d}z^2 +\frac{\partial u_1}{\partial \overline{z^2}}\,\mathrm{d}\overline{z^2}. $$ Thus, $\theta_+=\theta_-=\theta$, as claimed.

Note that this local normal form is unique up to replacement of $z^i$ by another holomorphic function of $z^i$ for $i=1,2$, while the choice of the two functions $u_i$ is arbitrary. It's not hard to show that, in dimension $4$, this Ansatz the only way that examples with equal Lee forms can arise. (É. Cartan would say that the local examples of bi-Hermitian metrics with equal Lee forms in dimension $4$ 'up to diffeomorphism' depend on two arbitrary functions of four variables.)

In higher dimensions, I expect that the analysis is more complicated, but, again, there will be too many examples to really classify because having equal Lee forms is a very underdetermined set of PDE on the data $(M,g,I_+,I_-)$.