For every Hilbert space $H$, $$K_0(\mathcal{K}(H))=\mathbb{Z}$$ because the only compact projections in $\mathcal{B}(H^{\oplus n})\cong M_n(\mathcal{B}(H))$ are finite-rank projections and these are distinguished only by dimension. Actually $$K_0(\mathcal{K}(E)) = \mathbb{Z}$$ for any Banach space $E$. Moreover, if $H$ is infinite-dimensional then $$K_0(\mathcal{B}(H))=\{0\},$$ because $H\cong H\oplus H$, hence $[{\rm id}_H]\oplus [{\rm id}_H] = [{\rm id}_H]$ so it must be the trivial group.

As Atkinson's theorem works well in non-separable spaces, the $K_1$-groups are the same as for the separable Hilbert space.