We will follow the OP's strategy and prove the following contrapositive form of the statement:
If a lattice is complemented and distributive, then every element of the lattice has a unique complement.
Convince yourself that this is equivalent to the claim in the question.
A complemented and distributive lattice is a boolean algebra, so we will use $+$ and $\cdot$ in place of $\vee$ and $\wedge$ respectively. Now, of course, every element does have a complement (by definition); the real task is to show uniqueness.
Let $x$ be an arbitrary element, and let $y$ and $z$ be its complements. We want to show that $y = z$. We start from $ y = y \cdot 1, $ and replace $1$ by $x+z$. Then applying distributivity and the fact that $yx = 0$, we get $ y = y(x+z) = yx + yz = 0 + yz = yz. \tag{1} $ Repeating this argument after switching $y$ and $z$, we get $ z = zy. \tag{2} $ Comparing $(1)$ and $(2)$, we are done.