Proving this modified classical 2D Ising Hamiltonian has a phase transition

Consider the following Hamiltonian with a local Hilbert space $mathcal{H}=mathcal{H}_Deltaotimes mathcal{H}_{Ising}congmathbb{C}^2otimesmathbb{C}^2 $. Denote an $Ltimes L $ square lattice as $Lambda(L)$
Let the classical 2D Ising model interactions with strength $J$ be denoted $$h_{Ising}^{i,j}(J)= J(|0rangle langle 0|^{(i)}otimes |1rangle langle 1|^{(j)} + |1rangle langle 1|^{(i)}otimes |0rangle langle0|^{(j)}). $$

Now define the local terms of a Hamiltonian on a 2D lattice be:
$$h^{i,j} = Delta(|0rangle langle 0|_{Delta}^{(i)}otimes |1rangle langle 1|_{Delta}^{(j)} + |1rangle langle 1|_{Delta}^{(i)}otimes |0rangle langle0|_{Delta}^{(j)} ) otimes mathbb{I}^{(i)}_{Ising}otimes mathbb{I}^{(j)}_{Ising} + |0rangle langle 0|_{Delta}^{(i)}otimes h_{Ising}^{i,j}(J) + |1rangle langle 1|_{Delta}^{(i)}otimes h_{Ising}^{i,j}(100J) + B |1rangle langle 1|_{Delta}^{(i)}otimes mathbb{I}_{Delta}^{(j)}otimes mathbb{I}_{Ising}^{(i)}otimes mathbb{I}_{Ising}^{(j)}.$$
We see this can be thought of as two spin-1/2 particles on each lattice site, such that one Ising model has interaction strength $Delta$ and an applied field $B$, and the other has strength either $J$ or $100J$ and has no applied field.
Assume that $Delta>>>J$, e.g. $Delta=10^{100}J$. Then we see that this model has multiple possible ground states depending on whether $B>0$ or $B<0$.

If $B>0$, then the ground state is one of $|0rangle^{Lambda(L)}|0rangle^{Lambda(L)}$ or $|0rangle^{Lambda(L)}|1rangle^{Lambda(L)}$. If $B<0$, then ground state is one of $|1rangle^{Lambda(L)}|1rangle^{Lambda(L)}$ or $|1rangle^{Lambda(L)}|1rangle^{Lambda(L)}$. Furthermore, since $Delta$ is so large, we see that all the low energy states are effectively the excited states of $H_{Ising}(J)=sum h^{i,j}_{Ising}(J)$ or $H_{Ising}(100J)=sum h^{i,j}_{Ising}(100J)$ depending on whether $B>0$ or $B<0$.

Since all the low energy states have these properties, then for $beta<<Delta$, we expect the model to act like $H_{Ising}(J)$ or $H_{Ising}(100J)$ depending on whether $B>0$ or $B<0$. Importantly, we expect a phase transition to occur close to the temperature at which the 2D Ising classical model undergoes a transition. As per Onsanger’s solution this is either at $$T approx frac{J}{k_B log(1+sqrt{2})}$$ or $$T approx frac{100J}{k_B log(1+sqrt{2})}.$$

I want to show that this is indeed the case, and that we expect a phase transition to occur here, corresponding to either the $H_{Ising}(J)$ or $H_{Ising}(100J)$ part of the Hamiltonian becoming paramagnetic. Since the zeros of the partition function indicate phase transitions, I wish to show that any zeros occur in the vicinity of $T_C$, and nowhere else.

My initial attempt was to expand the partition function in terms of the energy levels explicitly and show it was close to the 2D Ising mode partition function $Z(H_{Ising}(J))$, to get something like:
$$Z(H) = sum_{lambda_i} e^{-beta lambda_i(H)}
= sum_{lambda_i} e^{-beta lambda_i(H_{Ising}(J))} + e^{-4beta Delta} sum_{lambda_i} e^{-beta lambda_i(H’_{Ising}(J))} +e^{-6beta Delta} sum_{lambda_i} e^{-beta lambda_i(H”_{Ising}(J))} + dots
=Z(H_{Ising}(J)) + e^{-4beta Delta} sum_{lambda_i} e^{-beta lambda_i(H’_{Ising}(J))} +e^{-6beta Delta} sum_{lambda_i} e^{-beta lambda_i(H”_{Ising}(J))} + dots $$
where $H’_{Ising}(J))$ and $H”_{Ising}(J))$ are Ising models where one/two of the terms are 100J rather than J.

However, since the term $sum_{lambda_i} e^{-beta lambda_i(H’_{Ising}(J))}$ grows exponentially with the number of particles, then the $e^{-4beta Delta}$ is eventually not sufficient to supress those terms, and thus it is not clear $Z(H)approx Z(H_{Ising}(J)) + O(e^{-Delta beta})$.

Is there another convenient way to show that $Z(H)$ only has zeros near the critical temperature of the 2D Ising model? Alternatively, are there other methods to prove the transition will only occur there and nowhere else (e.g. non-analyticity in the order parameter)?