quantum mechanics - Why is it OK to keep the quadratic term in the small $hbar$ approximation?

I am following this^{[link broken]} set of notes:

Riccardo Rattazzi, The Path Integral approach to Quantum Mechanics, Lecture Notes for Quantum Mechanics IV, 2009, page 21.

I am having some issues to understand the small $\hbar$ expansion.

Consider the path integral in quantum mechanics giving the amplitude for a spinless particle to go from point $x_i$ to point $x_f$ in the time interval $T$ $$\int D[x]e^{i\frac{S[x]}{\hbar}}=\ldots$$ where $$S[x]=\int_{0}^{T}dt\,\mathcal{L}$$ let's assume now that the action has one stationary point $x_0$. Let's change the variable of integration in the path integral from $x$ to fluctuations around the stationary point $$x=x_0+y$$ $$\ldots=\int D[y]e^{i\frac{S[x_0+y]}{\hbar}}=\ldots$$ Let's Taylor expand the action around $x_0$ $$S[x_0+y]=S[x_0]+\frac{1}{2}\int_0^T dt_1dt_2\,\frac{\delta^2S}{\delta x(t_1)\delta x(t_2)}\bigg|_{x_0}y(t_1)y(t_2)+\ldots$$ which leaves us with $$\ldots=e^{i\frac{S[x_0]}{\hbar}}\int D[y]e^{\frac{i}{2\hbar}\int_0^T dt_1dt_2\,\frac{\delta^2S}{\delta x(t_1)\delta x(t_2)}\bigg|_{x_0}y(t_1)y(t_2)+\ldots}=\ldots \tag{1.65}$$ this is where the author considers the rescaling $$y=\sqrt{\hbar}\tilde{y}$$ which leaves us with $$\ldots=e^{i\frac{S[x_0]}{\hbar}}\int D[y]e^{\frac{i}{2}\int_0^T dt_1dt_2\,\frac{\delta^2S}{\delta x(t_1)\delta x(t_2)}\bigg|_{x_0}\tilde{y}(t_1)\tilde{y}(t_2)+\mathcal{O}(\hbar^{1/2})} \tag{1.66}$$ and we "obviously" have an expansion in $\hbar$, so when $\hbar$ is small we may keep the first term $$e^{i\frac{S[x_0]}{\hbar}}\int D[y]e^{\frac{i}{2}\int_0^T dt_1dt_2\,\frac{\delta^2S}{\delta x(t_1)\delta x(t_2)}\bigg|_{x_0}\tilde{y}(t_1)\tilde{y}(t_2)}$$ I do not like this rationale at all. It's all based on the rescaling of $y$ we have introduced, but had we done $$y=\frac{1}{\hbar^{500}}\tilde{y}$$ we wouldn't have obtained an expansion on powers of $\hbar$ on the exponent. What is the proper justification for keeping the quadratic term?