I'm going to use $\oplus$ to denote the xor operation.
Notice if $a \oplus b = c$. Then, xor'ing both sides, $$(a \oplus b) \oplus b = c \oplus b.$$
Applying a bit of associativity, $$a \oplus (b \oplus b) = c \oplus b.$$
By xor'ing something by itself, we always get $F$ (False), thus we have $$a \oplus F = c \oplus b.$$
By xor'ing with False, nothing happens so we just get $$c \oplus b = a$$
Thus, we can recover $a$ using $b$ and $c$. You can use a similar argument to show that $b$ is also recoverable using $a$ and $c$.
If you want to xor by more than two terms, then I believe you can use a combination of commutivity and associativity of xor to group the appropriate terms together and reduce it down to the two term case again.
I'm not quite sure if this property has a name.