Here, we will only consider the leading semi-classical approximation of a $1$-dimensional problem with Hamiltonian
$$ H(x,p) ~=~ \frac{p^2}{2m} + \Phi(x), $$
where $\Phi(x)<0$ is the potential function. Let us, for simplicity, assume that the potential $\Phi(x)=\Phi(-x)$ is an even function and strongly monotonically increasing for $x\geq 0$ with limit $\lim\limits_{|x|\to \infty}\Phi(x)=0$. Let $E_n<0$ denote the energy of the $n$'th bound state, ordered increasingly $E_1 <E_2<E_3<\ldots$, so that $E_1$ denotes the ground state energy. There is also a positive continuous unbounded spectrum $E\geq0$, which we shall not discuss further.
We shall here construct a counterexample of two potentials $\Phi^{\prime}(x)$ and $\Phi^{\prime\prime}(x)$ such that the quotient of potentials satisfies the limit
$$ \lim_{|x|\to \infty} \frac{\Phi^{\prime\prime}(x)}{\Phi^{\prime}(x)} ~=~1, \tag{1}\label{eq:1}$$
but where the corresponding quotient of bound state energies satisfies
$$ \lim_{n\to \infty} \frac{E_n^{\prime\prime}}{E_n^{\prime}} ~\neq~1. \tag{2}\label{eq:2}$$
The idea is to seek a potential $\Phi$ that would generate a bound state spectrum of the form
$$E_n= -R e^{-\mu n},\tag{3}\label{eq:3}$$
where $R>0$ is a Rydberg-like constant of dimension energy, and $\mu>0$ is a dimensionless positive constant. [The Hydrogen atom is for comparison $E_n= -R/n^2$.] Thus the number of states $N(E)$ below energy-level $E$ should roughly satisfy
$$E~\approx~ -R e^{-\mu N(E)} \qquad \Leftrightarrow \qquad N(E)~\approx~\frac{1}{\mu}\ln \left(-\frac{E}{R}\right). $$
This answer provides a semi-classical inversion formula for the potential $\Phi$ that we will use. The length $\ell(V)$ of the classically accessible region of the potential well at potential energy-level $V$ becomes
$$ 2\Phi^{-1}(V)~=~\ell(V) ~\approx ~\hbar\sqrt{\frac{2}{m}} \frac{\mathrm{d}}{\mathrm{d}V}\int_{V_{0}}^V \frac{N(E)~\mathrm{d}E}{\sqrt{V-E}} $$
$$~\approx~\frac{\hbar}{\mu}\sqrt{\frac{2}{m}} \left(\frac{2\arctan\sqrt{\frac{V_0}{V}-1}}{\sqrt{-V}} - \frac{\ln \left(-\frac{V_0}{R}\right)}{\sqrt{V-V_0}}\right). \tag{4}\label{eq:4} $$
One may check that the accessible length function $\ell(V)$ is a monotonically increasing function for $V\in[V_1,0[$ for some choice of the constants $V_0$ and $V_1$ with $V_0<V_1<0$. Asymptotically, such potential $\Phi(x)$ behaves as an inverse square potential $-C/x^2$ for $|x|\to\infty$, where $C>0$ is a positive constant.
We now construct the potential functions $\Phi^{\prime}(x)$ and $\Phi^{\prime\prime}(x)$ such that they are given by formula $\eqref{eq:4}$ in the outer region $x\geq x_1$, where $x_1:=\ell(V_1)/2>0$; and arbitrarily monotonically increasing in the inner region $0\leq x\leq x_1$. This implies that the quotient of potentials satisfies
$$ \frac{\Phi^{\prime\prime}(x)}{\Phi^{\prime}(x)} ~=~1 \qquad {\rm for}\qquad |x|\geq x_1, $$
so that condition $\eqref{eq:1}$ is satisfied.
For large enough states $n\geq n_1$, after the inner potential well is filled, the spectrum $E_n$ eventually becomes exponentially a la $\eqref{eq:3}$. We now choose the inner potentials $\Phi^{\prime}(x)$ and $\Phi^{\prime\prime}(x)$ such that there fits one more bound state into the profile $\Phi^{\prime\prime}(x)$ than $\Phi^{\prime}(x)$ for $|x|\leq x_1$. Then the labeling of states $E_{n+1}^{\prime\prime}\approx E_n^{\prime}$ would be off by one for $n\geq n_1$, yielding the inequality $\eqref{eq:2}$,
$$ \frac{E_n^{\prime\prime}}{E_n^{\prime}} ~\approx~e^{\mu} ~\neq~1. \qquad {\rm for}\qquad n\geq n_1.$$
