Being affine is not invariant under scissors relations. In other words, it is possible that $[X] = [Y]$ in the Grothendieck ring, where $X$ is affine and $Y$ is not.

For example, the diagonal $\Delta \subseteq \mathbf P^1 \times \mathbf P^1$ is ample, so $X = \mathbf P^1 \times \mathbf P^1 \setminus \Delta$ is affine. But
$$[X] = [\mathbf P^1 \times \mathbf P^1] - [\mathbf P^1] = (\mathbf L + 1)^2 - (\mathbf L + 1) = \mathbf L^2 + \mathbf L = [\mathbf P^2 - p],$$
and $\mathbf P^2 - p$ is not affine for any point $p \in \mathbf P^2$.

If you allow $k$-schemes with multiple components, every effective class has an affine representative. For example, a class of the form $[X]$ can be made affine by cutting $X$ into locally closed pieces that are affine (e.g. using Noetherian induction).

It could still be an interesting question which classes have a representative that is *irreducible* and affine (or irreducible smooth affine, or ...).