Let's think in terms of units. Density is mass per unit volume. Not only will different solutes have different masses dissolved into a unit volume, but the volume of the resulting solution will be different for different solutes.
When people say they can dissolve a unit of mass into a unit of volume (a common measure of solubility) they are not keeping track of the final volume of the resulting solution.
Here are the expressions for the density of the two saturated solutions (we should assume that the temperature and pressure are constant so that the density of the solvent itself is constant).
$\frac{m_{A,sat}}{V_{A,sat}} = \rho_{A,sat}$
$\frac{m_{B,sat}}{V_{B,sat}} = \rho_{B,sat}$
where $m$ and $V$ are masses and volumes, and $\rho$ is the density. $A$ and $B$ are the two solutes.
So if:
$\rho_{A,sat}{}\gt{} \rho_{B,sat}$
that could mean, in one limit where
$V_{A,sat} = V_{B,sat}$,
that
$m_{A,sat}{} \gt {}m_{B,sat}$.
Which is how I'm guessing you're thinking. If the volume change of solvation was similar for two solutes, one could trust that the density of the saturated solution with the (mass) solubility.
But there's another limit where there is a large difference in the volume change of solvation, between the two solutes. Then if:
$\rho_{A,sat}{}\gt{} \rho_{B,sat}$,
as before, and
$V_{A,sat}{} \lt{} V_{B,sat}$,
then it's possible that the unit mass dissolved in the two cases was equal,
$m_{A,sat} = m_{B,sat}$,
so that the two solubilities would be equal.
So no, with the above definitions of solubility and density, one cannot assume that the density of a saturated solution corresponds to the solubility.
