I don't know if this was the solution you were expecting. The residuals don't look too bad, especially for such a small amount of data.
xdata = [50 98 196 394 592]; %H2O2 initial concentration, uM
ydata = 1e6*[6.7848E-07 7.58E-07 6.55E-07 6.63E-07 6.39E-07
1.0936E-06 1.07E-06 1.13E-06 1.24E-06 1.21E-06
1.3570E-06 1.56E-06 1.75E-06 1.83E-06 2.02E-06
1.5725E-06 1.85E-06 2.35E-06 2.58E-06 2.78E-06
1.6762E-06 2.05E-06 2.55E-06 2.83E-06 3.04E-06]; %SCALED
ABTS = [1.01 2.01 4.02 6.04 8.05]';%TRANSPOSED
HRP = 2.00758;%SCALED
PP_fun = @(x,xdata) (x(1).*HRP)./(1+(x(2)./xdata)+(x(3)./ABTS));
lb = [0 0 0];
ub = [inf inf inf];
opts=optimoptions('lsqcurvefit', 'OptimalityTolerance',1e-12,...
'StepTolerance',1e-12, ...
'FunctionTolerance',1e-12,'Display','iter');
%Derive x0 by re-arranging as linear equations
A=cat(3, -HRP.*ones(size(ydata)), ydata./xdata, ydata./ABTS); A=reshape(A,[],3);
b=-ydata(:);
x0=(A\b)'
[x,resnorm,residual]= lsqcurvefit(PP_fun, x0,xdata, ydata,lb,ub,opts)
