= Solution
The <translated vector-space structure> follows by translating the new origin to the ordinary one using the <bijection> $T(\mathbf x)=\mathbf x-\mathbf a$. Transporting ordinary <scalar multiplication> through $T$ gives
$$
\boxed{k\odot\mathbf x=\mathbf a+k(\mathbf x-\mathbf a).}
$$
Then $T(\mathbf x\oplus\mathbf y)=T(\mathbf x)+T(\mathbf y)$ and $T(k\odot\mathbf x)=kT(\mathbf x)$, so the transported operations satisfy all <vector space> axioms. In particular, an explicit distributive check is
$$
k\odot(\mathbf x\oplus\mathbf y)=\mathbf a+k(\mathbf x+\mathbf y-2\mathbf a)
=(k\odot\mathbf x)\oplus(k\odot\mathbf y).
$$
The other distributive law follows in the same <coordinates>, or directly by expanding $(k+l)\odot\mathbf x$.
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