How do repulsive solute interactions affect the van't Hoff factor?
I think that your first probabalistic argument is correct, i.e that as the solute interaction increases the effective number of them is reduced. In the second case, as the interaction become repulsive, the effective number of them cannot increase over the maximum number there. They are restricted in their configurational entropy because the repulsion  means that the chance of finding one next to another is greatly reduced. Thus the probabilistic aligns with your stat-mech one.

You can calculate the entropy of mixing in an ideal solution then in a 'regular' one in which the components interact with one another.

In the ideal solution a statistical argument based on number of ways of placing particles on a grid,  leads to the entropy being given by

$$\Delta S_{mix}= -R[n_1\ln(x_1)+n_2\ln(x_2)]$$

for $n_1,\,n_2$ moles and $x_1,\,x_2$ mole fractions of species 1 and 2. 

In a solution in which there is interaction between (equal sized) molecules the free energy is 

$$\Delta G_{mix} = R[n_1\ln(x_1)+n_2\ln(x_2)] +(n_1+n_2)x_1x_2E$$ 

where E is is an energy of mixing. This can be calculated by considering that one solute molecule interacts with energy $\epsilon _{aa}$, the solvent with energy $\epsilon _{bs}$ and between solute and solvent with $\epsilon _{ab}$. Again a statistical argument based on molecules being on lattice sites means that nearest neighbour interactions of the type a-a  and b-b occur with probability $x_a^2$, and $x_b^2$ respectively and between a and b as $2x_ax_b$ and the 2 arise because a_b and b-a interactions are counted.

The total mixed energy is then $E_{tot} \approx (\epsilon_{aa}N_a^2+ 2\epsilon_{aa}N_aN_b+\epsilon_{bb}N_b^2 )$$

but the change on mixing is this energy less that due to the two components unmixed which is $\approx (N_a\epsilon_{aa}+N_b\epsilon_{bb})$

which means that the extra mixing energy is $E = zN_0(2\epsilon_{ab}-\epsilon_{aa}-\epsilon_{bb})$

where $z$ is related to the coordination number on the lattice. 

The entropy on mixing is then 

$$\Delta S_{mix}= -R[n_1\ln(x_1)+n_2\ln(x_2)] -(n_1+n_2)x_1x_2\frac{\partial w}{\partial T}$$

where the second term $\displaystyle \Delta S^E =-(n_1+n_2)x_1x_2\frac{\partial w}{\partial T}$ is a positive number about $0.2R$ J /mol/K for benzene/cyclohexane and $0.01R$ for benzene /$\ce{CCl4}$

Changing $\epsilon_{bb}$ as you suggest will still increase or decrease $\Delta S^E$ acting through the derivative.