Solution: Let us solve the given differential equation. \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = y(2-y) & = \frac{\mathrm{d} y}{y (2-y) } = \mathrm{d} x \\ & \implies \frac{1}{2} \left( \frac{1}{y} - \frac{1}{2-y} \right) \mathrm{d} y = \mathrm{d} x \\ & \implies \frac{1}{2} \left( \ln y - \ln (2-y) \right) = x + c \\ & \implies \ln \frac{y}{2-y} = 2 (x + c) \\ & \implies \frac{y}{2-y} = e^{2x + 2c} \\ & \implies \frac{y}{2-y} = C e^{2x},~ C = e^{2c} \\ & \implies y = 2 C e^{2x} - C y e^{2x} \\ & \implies y \left( 1 + C e^{2x} \right) = 2C e^{2x} \\ & \implies y = \frac{2 C e^{2x}}{1 + C e^{2x}}. \end{align*}

Now we will use the initial condition to see for what values of $a$ solution exists. \begin{align*} y(0) = a & \implies \frac{2C}{1 + C} = a \\ & \implies 2C = a + aC \\ & \implies C = \frac{a}{2 - a}. \end{align*} If $a = 2$, then solution does not exist, therefore the set $S$ will be $[0,\infty ) \setminus \{ 2 \} $ and hence the minimum of the set $S$ will be $0$.