Let me state a simple observation regarding the fibre of the 2-functor that arose in a conversation with Rui Loja Fernandes.

A symplectic groupoid $(\Sigma, \omega) \Rightarrow P$ induces a Poisson bivector $\pi$ on $P$, which is completely determined by the fact that the source and target maps are Poisson and anti-Poisson respectively (beware of the fact that **not** all Poisson structures arise in this fashion). Moreover, the foliation induced by the groupoid on $P$ corresponds to the symplectic foliation $\mathcal{F}$ of $\pi$. The symplectic groupoid contains all information regarding the Poisson geometry of $(P,\pi)$. A way to reformulate the question is to ask the following: given a Poisson manifold $(P,\pi)$ with symplectic groupoid $(\Sigma,\omega) \Rightarrow (P,\pi)$, what other Poisson structures on $P$ are there whose symplectic groupoid (viewed as a Lie groupoid) is $\Sigma \Rightarrow P$? Note that the requirement that the underlying Lie groupoid does not change fixes the symplectic foliation $\mathcal{F}$.

Fix a symplectic groupoid $(\Sigma,\omega) \Rightarrow (P,\pi)$. It is helpful to view $\pi$ as a *Dirac* structure on $P$ as follows. Set $\mathbb{T}P = \mathrm{T}P \oplus \mathrm{T}^*P$; consider

- the non-degenerate, symmetric bilinear product on $\mathbb{T}P$ defined by
$$ \langle (X,\alpha),(Y,\beta) \rangle = X(\beta) + Y(\alpha);$$
- the
*Courant bracket* $[.,.]_C:\Gamma(\mathbb{T}P) \times \Gamma(\mathbb{T}P) \to \Gamma(\mathbb{T}P)$ defined by
$$
[(X,\alpha ) , (Y,\beta)]_C = \left([X,Y], L_X \beta-L_{Y} \alpha + \frac{1}{2}\mathrm{d}(\alpha(Y) - \beta(X)\right),
$$
where $[.,.]$ denotes the standard Lie bracket and $L_X, L_Y$ denotes Lie derivatives.

A *Dirac* structure on $P$ is a subbundle $L \subset \mathbb{T}P$ such that

- $L$ is maximally isotropic with respect to the above bilinear product;
- $L$ is involutive under $[.,.]_C$.

Like Poisson bivectors, Dirac structures come equipped with a (possibly singular) foliation given by the projection onto the first factor of $\mathbb{T}P$.

The Poisson bivector $\pi$ defines a Dirac structure by

$$ L_{\pi} :=\{(\pi^{\sharp}(\alpha),\alpha) \,:\,\alpha \in \Omega^1(P) \}, $$

where $\pi^{\sharp} :\mathrm{T}^*P \to \mathrm{T}P$ is the bundle morphism defined by

$$ (\pi^{\sharp}(\alpha))(\beta) :=\pi(\alpha,\beta) $$

for $\alpha,\beta \in \Omega^1(P)$. The symplectic foliation $\mathcal{F}$ of $\pi$ coincides with the foliation associated to the Dirac structure.

Given a 2-form $B$ on $P$, define an endomorphism $\mathbb{T}P \to \mathbb{T}P$ by

$$ (X,\alpha) \mapsto (X,\alpha +B^{\flat}(X)), $$

where $(B^{\flat}(X))(Y) = B(X,Y)$. Denote the image of a Dirac structure $L$ under this endomorphism by $e^{B}L$; if $\mathrm{d}B=0$, then $e^B L$ is also a Dirac structure (this is known as a *gauge transformation* ). Note that gauge transformations do not change the underlying foliation associated to the initial Dirac structure $L$.

Let $L_{\pi}$ denote the Dirac structure associated to $\pi$ and, for a closed 2-form $B$ on $P$, consider $L' = e^B L_{\pi}$. For $L'$ to come from a bivector (which is necessarily Poisson), it is necessary that the endomorphism

$$ \mathrm{I} + B^{\flat} \circ \pi^{\sharp} : \mathrm{T}P \to \mathrm{T}P$$

be invertible. Under these conditions, obtain a new Poisson bivector $\pi'$ on $P$. It turns out that $(P,\pi')$ comes from a symplectic groupoid which, as a Lie groupoid, is isomorphic to the symplectic groupoid $(\Sigma,\omega) \Rightarrow (P,\pi)$. If $\omega'$ denotes the symplectic form on $\Sigma$ inducing the Poisson bivector $\pi'$, then $\omega' = \omega + \Omega$, where

$$\Omega = s^\ast B - t^\ast B $$

and $s,t :\Sigma \to P$ are the source and target maps respectively.

The above shows that the fibre of the 2-functor over $\Sigma \Rightarrow P$ that is considered in the question above contains the space of closed 2-forms $B$ on $P$ such that $\mathrm{I} + B^{\flat} \circ \pi^{\sharp}$ is invertible, where $\pi$ is a Poisson structure induced on $P$ by some multiplicative symplectic form on $\Sigma$. It remains to understand `how much more' there is to this fibre, but I have no good insights in this direction.