To me, the word "either" is unnecessarily confusing, and should be avoided if possible (of course, since it is part of the problem we have no choice in this case).
If "either $p$ or $q$" means the same thing as "$p$ or $q$", then the answer is simply $p\vee q$ (by the definition of $\vee$).
However, if "either $p$ or $q$" means "either $p$ or $q$, but not both" then this is equivalent to "$p$ is true and $q$ is false, or $p$ is false and $q$ is true". Do you see how to write the logical expression for this?
What we would like to prove is a conjunction, so it suffices to prove each conjunct separately and then glue them together at the end. This problem would probably be easier and more intuitive using proof by contradiction, but after talking with the asker, I will provide a direct proof.
$$\begin{array}{lr}
1. & (P \rightarrow Q)\wedge(Q \rightarrow R) & \text{Premise} \\
2. & P \rightarrow Q &\text{Simplification, 1}\\
3. & Q \rightarrow R & \text{Simplification, 1}\\
4. & \neg{P} \vee Q & \text{Conditional Law, 2}\\
5. & \neg{Q} \vee R & \text{Conditional Law, 3}\\
6. & Q \vee \neg{Q} & \text{Tautology} \\
7. & \neg{P} \vee R & \text{Constructive Dilemma, 4,5,6}\\
8. & P \rightarrow R & \text{Conditional Law, 7}\\
9. & (P \rightarrow Q) \vee (Q \rightarrow R) &\text{Addition, 2}\\
10. & (Q \rightarrow P) \vee (Q \rightarrow R) &\text{Addition, 3}\\
11. & (P \rightarrow Q) \vee (R \rightarrow Q) &\text{Addition, 2}\\
12. & Q \vee \neg{Q} & \text{Tautology}\\
13. & (Q \vee \neg{Q}) \vee (P \vee \neg{R}) & \text{Addidition, 12}\\
14. & (\neg{Q} \vee P) \vee (\neg{R} \vee Q) & \text{Associative Law, 13}\\
15. & (Q \rightarrow P) \vee (R \rightarrow Q) & \text{Conditional Law, 14}\\
16. & \big((P \rightarrow Q) \vee (Q \rightarrow R)\big)\wedge \big((Q \rightarrow P) \vee (Q \rightarrow R)\big) & \text{Conjunction, 9,10}\\
17. & \big((P \rightarrow Q) \vee (R \rightarrow Q)\big)\wedge \big((Q \rightarrow P) \vee (R \rightarrow Q)\big) & \text{Conjunction, 11,15}\\
18. & \big((P \rightarrow Q) \wedge (Q \rightarrow P)\big)\vee (Q \rightarrow R) & \text{Distributive Law, 16}\\
19. & \big((P \rightarrow Q) \wedge (Q \rightarrow P)\big)\vee (R \rightarrow Q) & \text{Distributive Law, 17}\\
20. & \Big(\big((P \rightarrow Q) \wedge (Q \rightarrow P)\big)\vee (Q \rightarrow R)\Big) \wedge & \\
&\Big(\big((P \rightarrow Q) \wedge (Q \rightarrow P)\big)\vee (R \rightarrow Q)\Big) & \text{Conjunction, 18,19}\\
21. & \big((P \rightarrow Q) \wedge (Q \rightarrow P)\big)\vee \big((Q \rightarrow R) \wedge (R \rightarrow Q) \big) & \text{Distributive Law, 20}\\
22. & (P \equiv Q) \vee (Q \equiv R) & \text{Definition of Biconditional, 21}\\
\therefore & (P \rightarrow R)\wedge \big((P \equiv Q) \vee (Q \equiv R)\big) & \text{Conjunction, 8,22}
\end{array}$$
As desired.
Best Answer
Assume left-to-right association for the $\star$. To satisfy $(p\star q)\star r \equiv s$ requires at least $3$ values for $p\star q$:
Then $p\star q$ can't be just true-or-false.