It seems unlikely that the Betti numbers can be determined by the data you list (bigraded Hilbert function and ranks of multiplication maps). Consider the following possibility: $\dim M_{1,0} = \dim M_{0,1} = 1$, $\dim M_{1,1} = 2$, and all other $M_{a,b}$ are zero; and the maps $y : M_{1,0} \to M_{1,1}$, $x : M_{0,1} \to M_{1,1}$ each have rank $1$ (all other multiplication maps are apparently zero).
Given this data and nothing else, it's impossible to tell if the maps from $M_{1,0}$ and $M_{0,1}$ to $M_{1,1}$ have the same image in $M_{1,1}$, or not.
An example where they have different images is $$M = (R/(x,y^2))[-1,0] \oplus (R/(x^2,y))[0,-1]$$ (where $R=k[x,y]$). An example where they have the same image is $$M' = (x,y)R/(x^2,y^2) \oplus k[-1,-1].$$ (The bracketed things are meant to be bidegree shifts, so $k[-1,-1]$ is supposed to be supported in bidegree $(1,1)$.)
$M$ is generated in bidegrees $(1,0)$ and $(0,1)$, while $M'$ requires generators in bidegrees $(1,0)$, $(0,1)$, and $(1,1)$. So, not even the first Betti degrees are determined by the data that you have.