symmetrical subtraction between $L_1 \in \mathsf{P}$ and $L_2 \in \mathsf{NPC}$
The answer is actually yes. Some intuition is that, when considering $\mathsf{NP}$-completeness (or in general, considering any complexity class under polynomial-time reductions), polynomial-time parts of the computation don't matter. Since $L_1$ is polynomial time, and $L_1 \Delta L_2$ and $L_2$ can be computed from each other knowing the answer to $L_1$, they should both be equivalent languages, up to polynomial-time computations.

Formally, to show "equivalence" we use polynomial time reductions. So you could solve this problem by:


Finding a polynomial-time reduction from $L_2$ to $L_1 \Delta L_2$. (How do we compute $L_1 \Delta L_2$ from $L_2$?)
Finding a polynomial-time reduction from $L_1 \Delta L_2$ to $L_2$. Hint: use the fact that
$$(L_1 \Delta L_2) \Delta L_1 = L_2$$


Once you have done this, you have shown that $L_2$ and $L_1 \Delta L_2$ are equivalent; if you are familiar with the notation,
$$
L_2 \le_p L_1 \Delta L_2 \le_p L_2
$$
which  means if one of them is $\mathsf{NP}$-complete, then both of them are.