There are no other non-entangling gates in $SU(d^2)$ in any dimension $d=2,3,\dots$. Note that the global phase is irrelevant to the problem, so we lose no generality by considering non-entangling gates in $U(d^2)$ instead. We will prove that if $U\in U(d^2)$ is non-entangling then either $U\in U(d)\otimes U(d)$ or $\text{SWAP}\circ U\in U(d)\otimes U(d)$.
Preliminaries
If $A$ is a linear subspace of $\mathbb{C}^d$ and $|\psi\rangle\in\mathbb{C}^d$ a pure state, then let $|\psi\rangle\otimes A$ denote the set $\{|\psi\rangle\otimes|\phi\rangle:|\phi\rangle\in A\}$ which is a linear subspace of $\mathbb{C}^d\otimes\mathbb{C}^d$. Similarly, for $A\otimes|\phi\rangle$. We say that a linear space $B\subseteq\mathbb{C}^d\otimes\mathbb{C}^d$ is entanglement-free if every element of $B$ is (a scalar multiple of) a product state.
Lemma If a linear subspace $B\subseteq\mathbb{C}^d\otimes\mathbb{C}^d$ is entanglement-free, then either $B=A\otimes|\psi\rangle$ or $B=|\psi\rangle\otimes A$ for some state $|\psi\rangle\in\mathbb{C}^d$ and some linear subspace $A\subseteq\mathbb{C}^d$.
Proof. Assume otherwise. Then we can find $|a\rangle\otimes|b\rangle\in B$ and $|x\rangle\otimes|y\rangle\in B$ such that $|x\rangle$ is not a scalar multiple of $|a\rangle$ and $|y\rangle$ is not a scalar multiple of $|b\rangle$. However, then $|a\rangle\otimes|b\rangle + |x\rangle\otimes|y\rangle$ is entangled$^1$.$\square$
Two types of non-entangling gates
Now, suppose that $U\in U(d^2)$ is non-entangling. Then the image $U[B]$ of any entanglement-free subspace $B=A\otimes|\psi\rangle$ under $U$ is entanglement-free and the lemma above implies that either
$$
U[A\otimes|\psi\rangle] = A'\otimes|\psi'\rangle\tag{1}
$$
or
$$
U[A\otimes|\psi\rangle] = |\psi'\rangle\otimes A'\tag{2}
$$
for some subspace $A'$ of $\mathbb{C}^d$ and some state $|\psi'\rangle\in\mathbb{C}^d$. In the first case, $U\in U(d)\otimes U(d)$. In the latter case $\text{SWAP}\circ U\in U(d)\otimes U(d)$. Since $(1)$ and $(2)$ exhaust all possibilities, no other non-entangling gates exist.
Intuition
The argument above attempts to capture the intuitive observation that if we vary$^2$ the state of the first qudit in a product state that is fed into a two-qudit unitary gate then that variation affects either the first qudit, the second qudit or both qudits at the output. However, if the variation affects both qudits then they become entangled$^3$. Therefore, since the gate is non-entangling, the variation can only feed through either to the first qudit or to the second qudit. These two cases correspond to the two possibilities $(1)$ and $(2)$ above.
Normalizer
The normalizer $N:=N(U(d)\otimes U(d))$ of $U(d)\otimes U(d)$ in $U(d^2)$ does indeed have two connected components which correspond to the identity and the SWAP gate. First, note that every non-entangling gate belongs to $N$. Conversely$^4$, no entangling gate belongs to $N$.
Now, $N$ inherits its topology from $U(d^2)$ which inherits its topology from $\mathbb{C}^{d^4}$. Moreover, $N$ is closed, so connectedness and path-connectedness are equivalent in $N$. Thus, if $N$ was connected, then there would be a continuous path from a gate of the form $U_1\otimes V_1$ to a gate of the form $\text{SWAP}\circ(U_2\otimes V_2)$. However, this would mean that we can approximate the SWAP gate by product gates arbitrarily well, which is impossible. Therefore, $N$ has at least two connected components.
Finally, $U(d)$ is path-connected, so we can form a continuous path between any two gates. Taking the product of such paths, we see that any two gates of the form $U_1\otimes V_1$ live in the same connected component. Similarly for gates of the form $\text{SWAP}\circ(U_2\otimes V_2)$. Therefore, $N(U(d)\otimes U(d))$ has exactly two connected components.
$^1$ This can be proved rigorously by extending $\{|a\rangle\otimes|b\rangle\}$ to a basis and writing the coefficients of $|a\rangle\otimes|b\rangle + |x\rangle\otimes|y\rangle$ in that basis as a $d\times d$ matrix. Since $|x\rangle$ is not a scalar multiple of $|a\rangle$ and $|y\rangle$ is not a scalar multiple of $|b\rangle$, the matrix is full rank and therefore $|a\rangle\otimes|b\rangle + |x\rangle\otimes|y\rangle$ is not a product state.
$^2$ For example, we could imagine varying the state of the first qudit with time as in $|\psi(t)\rangle\otimes|\phi\rangle$.
$^3$ More generally, the qudits could become correlated classically. However, this possibility is ruled out by unitarity. It would be relevant if we considered two-qudit quantum channels instead of two-qudit unitary gates.
$^4$ We can arrange for the conjugation of a product unitary with non-degenerate spectrum by an entangling unitary to result in an operator with entangled eigenstates. Such an operator is not a product unitary.
