Based on the description given in the question, we can build the following truth table: \begin{matrix} A & A & | & ⊕ \\ \hline F & F & | & F & \color{red}{\text{A ⊕ A}}\\ F & T & | & T & \color{blue}{\text{¬A ⊕ A}}\\ T & F & | & T & \color{blue}{\text{A ⊕ ¬A}}\\ T & T & | & F & \color{red}{\text{A ⊕ A}} \\ \end{matrix} Now, we can deduce $⊕$ is the exclusive or operation. Deduced directly from the rules stated in the question.
Addednum: We can easily deduce the meaning of $\neg.$ Given the set $\mathbb{B} = \{ T, F \},$ any unary operator $\neg : \mathbb{B} \to \mathbb{B}$ will either operate as identity or as negation. Since the 2 equations in the given problem differ only by $\neg,$ $ A \oplus \neg A = T, A \oplus A = F, $ we can deduce that $\neg$ can not be identity (assume $\neg A \equiv A$ and you'll get contradiction in the given system of formulas). Hence $\neg$ is negation. QED. Now we proceed as above.