Yes, SWAP is a Clifford gate.
Your proof is correct. By definition, an $n$-qubit gate $U$ is Clifford if $UPU^\dagger\in G_n$ for all $P\in G_n$ where $G_n$ is the $n$-qubit Pauli group. However, it is easy to see that we only need to check that $UQU^\dagger\in G_n$ for $Q$ that are generators of $G_n$. Moreover, $G_n$ is generated by $\{i, Z_k, X_k\,|\,k=1,\dots,n\}$ where $X_k$ denotes the tensor product of Pauli $X$ on the $k$th qubit and identity applied to all other qubits and similarly for $Z_k$. Since $UiU^\dagger=iI\in G_n$ for all $U$, we only have to check that $UX_kU^\dagger\in G_n$ and $UZ_kU^\dagger\in G_n$ for $k=1,\dots,n$. This is exactly what the table in the question accomplishes.
It is easy to check that
$$
\text{SWAP} = C_1NOT_2 \circ C_2NOT_1 \circ C_1NOT_2\tag1
$$
where $C_iNOT_j$ denotes the CNOT gate with qubit $i$ as control and $j$ as the target, e.g. by applying both sides of $(1)$ to the computational basis states.