A simple model that exhibits emergent symmetry?

The simplest model is the spin-1/2 chain with Majumdar–Ghosh interaction: $$H=\sum_i P_{3/2}(i-1,i,i+1),$$ where $P_{3/2}(i,j,k)$ is the projection operator that projects a state onto the subspace with total spin-3/2 on sites $i,j,k$. The ground states are two dimer states (see the figure on wikipedia Majumdar–Ghosh model): $$|\psi_1\rangle=\prod_i|\mathrm{singlet}\rangle_{2i,2i+1},$$ $$|\psi_2\rangle=\prod_i|\mathrm{singlet}\rangle_{2i-1,2i}.$$

If we define the symmetry transformation $U(i,j)=\exp(ia_{ij}P_0(i,j))$ where $P_0(i,j)$ is the singlet projection operator, then $$U(2i,2i+1)|\psi_1\rangle=\exp(i a_{2i,2i+1})|\psi_1\rangle,$$ $$U(2i-1,2i)|\psi_2\rangle=\exp(i a_{2i-1,2i})|\psi_2\rangle,$$ for any $i$. In other words, $|\psi_1\rangle$ supports a one dimensional representation of the group $U(2i,2i+1)$ (any $i$) which is not a symmetry of the original Hamiltonian. Similar for $|\psi_2\rangle$. It is exactly those emergent symmetries that make this model soluble.

More sophisticated examples can be found here: 0207106.

I think the simplest example is very closely related to your suggestion of the two-site Ising model. Instead, consider the two-site XX-chain: $$ H = \sigma_1^x \sigma_2^x + \sigma_1^y \sigma_2^y. $$ Clearly the Hamiltonian has $U(1)$ symmetry (generated by $\sigma^z_1 + \sigma^z_2$) but it does NOT have full $SU(2)$ symmetry. However, its (unique!) ground state is the spin singlet $$ |\psi_\textrm{gs}\rangle = \frac{1}{\sqrt{2}} \left( |\uparrow_1 \downarrow_2\rangle -|\downarrow_1 \uparrow_2\rangle \right). $$ Hence, its unique ground state has an emergent $SU(2)$ symmetry.