Cancellation Law for Direct Sums - What is wrong with this argument?

All that is going on is that $G\oplus H\cong G\oplus K$ does not imply $(G\oplus H)/G\cong(G\oplus K)/G$. This should not be surprising, because the quotient $(G\oplus H)/G$ depends not only on the group $G\oplus H$ but also on the choice of a subgroup $G$, and the isomorphism $G\oplus H\cong G\oplus K$ is not assumed to preserve the choice of subgroup. If $\varphi:G\oplus H\to G\oplus K$ is an isomorphism, it gives an isomorphism $(G\oplus H)/G\cong (G\oplus K)/\varphi(G)$, since cosets of $G$ in $G\oplus H$ map to cosets of $\varphi(G)$ in $G\oplus K$. But the image $\varphi(G)$ may not be the same as the subgroup $G$ of $G\oplus K$, so this doesn't tell you anything about $(G\oplus K)/G$.

which just happens to be isomorphic to the original group.

Yes, this is is what happens, and also what underlies your original example. The statement $G \oplus H \cong G \oplus K \implies H \cong K$ can be wrong because the factor $G$ can appear in "different ways" on each side.

Consider the stronger assumption: not only does $G \oplus H \cong G \oplus K$, but there is an isomorphism $f: G \oplus H \to G \oplus K$ such that for each $g \in G$ we have $f((g,0)) = (g, 0)$. That is, the isomorphism "leaves $G$ intact", so that $G$ "appears the same way on both sides". Then if we compose $f$ with the projection $\pi_K : G \oplus K \to K$, we obtain that the $\ker(\pi_K \circ f) = G \oplus \{0\}$, and so $K \cong (G \oplus H)/(G \oplus \{0\}) \cong H$ by the first isomorphism theorem.