$A \oplus B = A \oplus C$ imply $B = C$? I don't quite yet understand how $\oplus$ (xor) works yet. I know that fundamentally in terms of truth tables it means only 1 value(p or q) can be true, but not both.
But when it comes to solving problems with them or proving equalities I have no idea how to use $\oplus$.
For example: I'm trying to do a problem in which I have to prove or disprove with a counterexample whether or not $A \oplus B = A \oplus C$  implies $B = C$ is true. 
I know that the venn diagram of $\oplus$ in this case includes the regions of A and B excluding the areas they overlap. And similarly it includes regions of A and C but not the areas they overlap. It would look something like this:

I feel the statement above would be true just by looking at the venn diagram since the area ABC is included in the $\oplus$, but I'm not sure if that's an adequate enough proof. 
On the other hand, I could be completely wrong about my reasoning.  
Also just for clarity's sake: Would $A\cup B = A \cup C$ and $A \cap B = A \cap C$ be proven in a similar way to show whether or not the conditions imply $B = C$? A counterexample/ proof of this would be appreciated as well.

This can be done using a simple calculation.  But I don't know how to interpret your question, so I'll give two answers. :-)

I'm assuming $\;A,B,C\;$ are booleans.  I will write $\;\not\equiv\;$ instead of $\;\oplus\;$, and $\;\equiv\;$ instead of $\;=\;$ on booleans.
First, note that $\;\equiv\;$ and $\;\not\equiv\;$ are not only both associative, but they are also mutually associative.  Therefore no parentheses are needed in the following calculation.
We can now simplify $\;A \oplus B = A \oplus C\;$ as follows:
\begin{align}
& A \not\equiv B \equiv A \not\equiv C \\
\equiv & \;\;\;\;\;\text{"rearrange"} \\
& A \not\equiv A \equiv B \not\equiv C \\
\equiv & \;\;\;\;\;\text{"simplify"} \\
& \text{false} \equiv B \not\equiv C \\
\equiv & \;\;\;\;\;\text{"simplify"} \\
& B \equiv C \\
\end{align}

If instead $\;A,B,C\;$ are sets, and your $\;\oplus\;$ is the symmetric difference of two sets (which is normally written as $\;\triangle\;$), then the proof is slightly longer, but with essentially the same structure.
The simplest definition of symmetric difference is
$$
x \in A \oplus B \equiv x \in A \not\equiv x \in B
$$
We can expand the definitions and simplify using logic, as follows:
\begin{align}
& A \oplus B = A \oplus C \\
\equiv & \;\;\;\;\;\text{"set extensionality; definition of $\;\oplus\;$, twice"} \\
& \langle \forall x :: x \in A \not\equiv x \in B \equiv x \in A \not\equiv x \in C \rangle \\
\equiv & \;\;\;\;\;\text{"logic: rearrange"} \\
& \langle \forall x :: x \in A \not\equiv x \in A \equiv x \in B \not\equiv x \in C \rangle \\
\equiv & \;\;\;\;\;\text{"logic: simplify"} \\
& \langle \forall x :: \text{false} \equiv x \in B \not\equiv x \in C \rangle \\
\equiv & \;\;\;\;\;\text{"logic: simplify"} \\
& \langle \forall x :: x \in B \equiv x \in C \rangle \\
\equiv & \;\;\;\;\;\text{"set extensionality"} \\
& B = C \\
\end{align}

In both cases, we have found a stronger conclusion than was asked: we proved equivalence of the two expressions.
