Does a shift operation distribute over XOR
Shift is right-distributive over XOR, and over many other operations as well (AND, OR, addition, min, max)

As for why, informally for the bitwise operations I'd explain it as: bitwise operations are like pointwise operations (if you view a number as a function from bit index to the bit at that index), translating (or permuting, etc) the positions distributes over doing the operation. Eg if we have two numbers $f$ and $g$ as such functions, then $(f ⊕ g)(k) = f(k) ⊕ g(k)$ so $(f ⊕ g)(k + j) = f(k + j) ⊕ g(k + j)$, here $f(k + j)$ is just $f$ shifted by $j$ so this showed that it didn't matter whether the shift is done after the XOR or on both of its operands. In general this applies to any remapping of the coordinates, ie $(f ⊕ g)(r(k)) = f(r(k)) ⊕ g(r(k))$ where $r$ is any function from $\mathbb{N}$ to $\mathbb{N}$.

As pointed out in the comments, this also relies on the operator to map two zeroes to an other zero, ie $0 ⊕ 0 = 0$.

Or simpler but less general, using string notation, let's say we have two number $a$ and $b$ and let their XOR be $c$, then $a0^k ⊕ b0^k = c0^k$ and $(a⊕b)0^k = c0^k$ (by substitution).