How does reduction of a decision problem work? I am given the following problem description:

Given $l$ lists, $L_1$, $L_2$, . . .$L_l$ each containing $N$ bit vectors of $n$ bits each, we want to find tuples $(x_1,···,x_l)$ with $x_i$ in the corresponding $L_i$, such that:$$⊕_{i=1}^l x_i = S $$ for some target n-bit vector $S$.  In other words, we want to find one element from list such that the XOR of the chosen elements is the target $S$

Now I am required to show that if we can solve the equation in the case where $S=0$, we can for the same cost solve it for an arbitrary value (with exact same parameters, $N$, $n$, and $l$).  I am given a hint to proceed by reducing an instance with $S \neq 0$ to an instance with $S=0$.
What am I missing that I am not able to reduce it?
Let $I$ be your instance and consider the instance $I'$ obtained by replacing each bit vector $y$ in $L_1$ with $y' = y \oplus S$ (where $\oplus$ denotes bitwise xor). Call $L'_1$ the list containing all vectors $y'$.
Consider the tuples $(x_1, x_2, \dots, x_l)$ and $(x'_1, x_2, \dots, x_l)$ where $x_i \in L_i$ and $x'_1 \in L'_1$.
We have:
$$
      x_1 \oplus ( \oplus_{i=2}^l x_i ) = \oplus_{i=1}^l x_i  = S \iff  (x_1 \oplus S) \oplus  ( \oplus_{i=2}^l x_i ) = x'_1\oplus  ( \oplus_{i=2}^l x_i ) = S \oplus S = 0,
$$
showing that there is a solution to $I$ iff there is a solution to $I'$ (and you can convert one solution into the other in linear time).