There are many functions and methods for solving a set of equations. These can't be exactly compared to each other. Some are slower and have special capabilities. Some are faster and more likely to return worse answers. FindInstance and excel's solver are very different things and would be used in very different situations.

Here is how I would go about looking for a quick answer to the problem assuming that the solution is somwhere near the starting values ({{dr, 1}, {ds, 1}, {ph, 1}, {pl, 1}, {yr, 1}, {ys, 1}, {[Rho], 1}})

equations = {((pi*(1 - \[Lambda]h)*(1 - ph))/(ys + 
        ph*(1 - ys) - \[Lambda]h*
         ds)) + (((1 - pi)*(1 - \[Lambda]l)*(1 - pl))/(ys + 
        pl*(1 - ys) - \[Lambda]l*ds)) == 
   0, (pi*\[Lambda]h*((1/
          ds) - (1 - \[Lambda]h)/(ys + 
           ph*(1 - ys) - \[Lambda]h*ds)) + (1 - 
        pi)*\[Lambda]l*((1/
          ds) - (1 - \[Lambda]l)/(ys + 
           pl*(1 - ys) - \[Lambda]l*ds))) == 
   0, ((pi*(1 - ph))/(yr + 
        ph*(1 - yr))) + (((1 - pi)*(1 - \[Lambda]l)*(1 - pl))/(yr + 
        pl*(1 - yr) - \[Lambda]l*dr)) == 
   0, (1 - pi)*\[Lambda]l*((1/
        dr) - (1 - \[Lambda]l)/(yr + pl*(1 - yr) - \[Lambda]l*dr)) == 
   0, \[Rho]*(ys - \[Lambda]l*ds) == (1 - \[Rho])*(\[Lambda]l*dr - 
       yr) && \[Rho]*(ys - \[Lambda]h*ds)/ph == (1 - \[Rho])*(1 - yr),
   pi*(\[Lambda]h*Log[ds] + (1 - \[Lambda]h)*
        Log[((ys + 
             ph*(1 - ys) - \[Lambda]h*
              ds)/((1 - \[Lambda]h)*(ph/R)))]) + (1 - 
       pi)*(\[Lambda]l*Log[ds] + (1 - \[Lambda]l)*
        Log[((ys + 
             pl*(1 - ys) - \[Lambda]l*
              ds)/((1 - \[Lambda]l)*(pl/R)))]) == 
   pi*(\[Lambda]h*Log[(yr + ph*(1 - yr))] + (1 - \[Lambda]h)*
        Log[(yr + ph*(1 - yr))]) + (1 - 
       pi)*(\[Lambda]l*Log[dr] + (1 - \[Lambda]l)*
        Log[((yr + 
             pl*(1 - yr) - \[Lambda]l*
              dr)/((1 - \[Lambda]l)*(pl/R)))])}

FindRoot[equations, {{dr, 1}, {ds, 1}, {ph, 1}, {pl, 1}, {yr, 1}, {ys, 1}, {\[Rho], 1}}]