If you write out the truth tables then it's pretty simple, write the truth table for A=C⊕B, you can chose A, B or C to be the output and you get a xor truth table with the remaining two as inputs, so if A=C⊕B then B=C⊕A and C=A⊕B
But I was wondering if there was a nicer proof using Boolean algebra only
Edit: A=C⊕B can be written as A=(C'B)+(CB')
$A=C \oplus B$
$\oplus A$ both sides. (Yes, you can do that).
$$A\oplus A = A\oplus C \oplus B$$
$$\implies FALSE = A\oplus C \oplus B$$
Now, $\oplus B$ both sides.
$$B \oplus FALSE = A\oplus C \oplus B \oplus B$$
(We are also using commutativity of $\oplus$ above)
$$B \oplus FALSE = A\oplus C \oplus FALSE$$
$$\implies B = A \oplus C$$.
Repeat similar construction for $C$.