I figured a positive eigenvalue. For $\lambda>0$, if $f(x) := \alpha\exp(\tau x)+ \exp(-\tau x)$ with $\alpha$ a constant and $\tau=\sqrt{2/\lambda}$ then $Af-\lambda f=0$ if and only if $$\frac{e^{-\tau}-\alpha e^{\tau}-(\alpha-1)}{\tau}x+\frac{(-1-\tau) e^{-\tau}-\alpha(1-\tau)e^{\tau}-\alpha-1}{\tau^2}=0,\text{ for all }0\le x\le1.$$ To annihilate the linear term in x, you can solve for $\alpha$ in terms of $\tau$ to obtain $$\alpha=\frac{1+e^{-\tau}}{1+e^{\tau}}.$$ Substituting this value on the other term gives $$\frac{(-1-\tau) e^{-\tau}-\alpha(1-\tau)e^{\tau}-\alpha-1}{\tau^2}=\frac{(-\tau-2) e^{-\tau}-4+(\tau-2)e^{\tau}}{1+e^\tau},$$ which vanishes only when $$e^\tau=\frac{\tau+2}{\tau-2},$$ and a plot suggests that this is possible at a unique value of $\tau\approx 2.4$.
So there is a unique positive eigenvalue, which happens to be simple!