Establishing transfer function for s = f(e)
I think I found the answer, I'll give the example of D1 closed (not blocked) and D2 open (blocked).

As can be seen in the image below, a closed diode behaves like a voltage source of Vt Volts while a blocked diode behaves like an open switch:



Exercice Data:
$$
V_T = 0.6V \\
E_1 = E_2 = 1V
$$

Using KVL we have:
$$
s = e - V{_d{_1}} - E_1 \\
I = I_1 = (e - V{_d{_1}} - E_1)/R \\
$$

And the fact that D1 is closed and D2 is open means that:
$$
V{_d{_1}} = V{_T} \rightarrow I_1 &gt; 0 \\
I_2 = 0 \rightarrow V{_d{_2}} &lt; V_T           
$$

Replacing Vd1, Vd2 and I1 with their formulas we get:
$$
e - s - E1 = V_T \rightarrow (e - V{_d{_1}} - E_1)/R &gt; 0 \\
I_2 = 0 \rightarrow s - e - E_2 &lt; V_T
$$

In the end that leaves us with:
$$
s = e - E_1 - V_T \rightarrow e &gt; V{_d{_1}} + E_1 \\
I_2 = 0 \rightarrow s &lt; e + E_2 + V_t
$$

The result is given to us by the numerical application:
$$
e &gt; 1.6V \\
s = e - 1.6V
$$

This should be correct and all the other cases are similar. If this is not correct, please notify.