The statement that one may add a $Q$ exact term to the action $S\to S+Q\chi$ without altering the correlators is not always strictly true. One has to be careful to make a choice of $\chi$ that does not change the asymptotic behaviour of the action $S$ at the boundary of field space. The choice of $\chi=-\psi$ is clearly a choice that drastically alters the asymptotic behaviour of $S$.
Furthermore, $\left<Q\mathcal{O}\right>$ only vanishes up to boundary terms in field space. Non-zero boundary terms would cause the standard arguments you made to fail.
Edit:
This can be demonstrated by a finite dimensional example which can be found in section 3.22 of the lectures by G. Moore on Donaldson Invariants and 4-manifolds at: http://www.physics.rutgers.edu/~gmoore/SCGP-FourManifoldsNotes-2017.pdf.
Consider the supersymmetric integral $$\mathcal{Z}=\int d\mathcal{M}e^{-S},\quad S=\frac{1}{2}H^2+iHs(x)-i\bar{\psi}\frac{ds(x)}{dx}\psi,\quad d\mathcal{M}=\frac{dxdHd\psi d\bar{\psi}}{2\pi i},$$
where $s:\mathbb{R}\to\mathbb{R}$ is a real valued function satisfying $|s(x)|\to\infty$ as $|x|\to\infty$. The action is invariant under a supersymmetry with $$Qx=\psi,\quad Q\psi=0,\quad Q\bar{\psi}=H,\quad QH=0.$$ Furthermore, the action is $Q$-exact $$S=Q\Psi,\quad \Psi=\left(\frac{1}{2}\bar{\psi}H+i\bar{\psi}s(x)\right).$$
Upon integrating out $\psi,\bar{\psi}$ and the auxiliary $H$ we find that $\mathcal{Z}$ is given by
$$\mathcal{Z}=\int^{\infty}_{-\infty}\frac{dx}{\sqrt{2\pi}}s'(x)e^{-\frac{1}{2}s(x)^2}=\deg(f)\int^{\infty}_{-\infty}\frac{dy}{\sqrt{2\pi}}e^{-\frac{1}{2}y^2}=\deg(f)=\sum_{z(s)}\frac{s'(x_0)}{|s'(x_0)|}$$
to evaluate the the integral we changed variables $f:x\mapsto y=s$ and $\deg(f)$ denotes the degree of that map, finally $z(s)=\{x_0|s(x_0)=0\}$ is the zero set of $s$. Note that upon integrating out $H$, i.e. setting $H=-is(x)$, $z(s)$ coincides precisely with the set of $Q$ fixed points.
However, we could have considered deforming the action by a $Q$ exact term, say, $S\to S+Q(i\bar{\psi}t(x))$ and repeating the above steps yields that $$\mathcal{Z}_t=\int^{\infty}_{-\infty}\frac{dx}{\sqrt{2\pi}}(s'(x)+t'(x))e^{-\frac{1}{2}(s(x)+t(x))^2}\stackrel{?}{=}\mathcal{Z}.$$
It should be clear that there are choices of $t$ such that $\mathcal{Z}_t$ doesn't exist, we require a $t$ such that $|s(x)+t(x)|\to\infty$ as $|x|\to\infty$ still holds. Even if this holds true there are choices such that $\mathcal{Z}_t\neq\mathcal{Z}$, as a simple demonstration you could just choose $s(x)=x^2-2x+1$ and $t(x)=-x^2$ then $z(s)=\{1\}$ and $s'(1)=0$ hence $\mathcal{Z}=0$ on the other hand the zero locus $z(s+t)=\{1/2\}$ $s'(1/2)+t'(1/2)=-2$ hence $\mathcal{Z}_{t=-x^2}=-1$.
