If $k$ is an algebraically closed field of characteristic $0$, then the map

$$X \mapsto N(X) := \#(\text{connected components of }X)$$

defined for smooth projective varieties extends to a ring homomorphism $K_0(Var/k) \to \mathbf{Z}$. So if both $X$ and $Y$ are smooth projective, then $[X] + [Y] = 0$ implies that $N(X) = N(Y) = 0$, hence $X$ and $Y$ are both empty.

If $X$ or $Y$ is not necessarily smooth projective, then the above invariant is not strong enough and we have to use the Poincaré polynomials (or the Hodge polynomials as SashaP suggested in his comment in Artur Jackson's answer). Here is the outline.

Again, $k$ is an algebraically closed field of characteristic $0$. Fix a Weil cohomology theory $H^\bullet$,
then the map defined for smooth projective varieties

$$X \mapsto P(X;t) := \sum_{i}(-1)^ib_{i}(X)t^i $$
where $b_i(X) := \dim H^i(X)$,
extends to a ring homomorphism $K_0(Var/k) \to \mathbf{Z}[t]$. One checks that for any variety $X$ (not necessarily smooth nor projective), the associated Poincaré polynomial $P(X;t)$ is of degree $2 \dim X$ (whose leading coeeficient is the number of irreducible components of $X$). Accordingly if $[X] = 0$ for some variety $X$, then $X$ is empty. In particular if $[X \sqcup Y] = [X] + [Y] = 0$, then both $X$ and $Y$ are empty.