This is more of an extended comment, rather than an answer. Consider the Lie trivial groupoid of the form $M\Rightarrow M$, where $M$ is a closed oriented surface. Then any two symplectic forms on $M$ with the same total integral over $M$ are symplectomorphic (Moser deformation argument). Hence the volume of your area form is a symplectic invariant. So the fiber of your map contains the category is equivalent to the set $\mathbb{R} \smallsetminus {0}$. But there is more: take a covering space $\tilde{M}$ of $M$. The fundamental group $\pi_1 (M)$ acts on $\tilde{M}$ and the action groupoid $\pi_1 (M)\times \tilde{M}\Rightarrow \tilde{M}$ is Morita equivalent to $M$. It is also a symplectic groupoid in your fiber for any choice of a $\pi_1(M)$-invariant symplectic form on $\tilde{M}$...
(edit) Unfortunately the form on $\pi_1 (M)\times \tilde{M}\Rightarrow \tilde{M}$ does not look multiplicative, as Daniele Sepe points out. Oops.
This is more of an extended comment, rather than an answer. Consider the Lie trivial groupoid of the form $M\Rightarrow M$, where $M$ is a closed oriented surface. Then any two symplectic forms on $M$ with the same total integral over $M$ are symplectomorphic (Moser deformation argument). Hence the volume of your area form is a symplectic invariant. So the fiber of your map contains the category is equivalent to the set $\mathbb{R} \smallsetminus {0}$. But there is more: take a covering space $\tilde{M}$ of $M$. The fundamental group $\pi_1 (M)$ acts on $\tilde{M}$ and the action groupoid $\pi_1 (M)\times \tilde{M}\Rightarrow \tilde{M}$ is Morita equivalent to $M$. It is also a symplectic groupoid in your fiber for any choice of a $\pi_1(M)$-invariant symplectic form on $\tilde{M}$...