Here I'm taking the Yamabe invariant of $(M,g)$ to mean the conformal invariant $Y(g)$ as given here, not the supremum $\sigma(M)$ over all metrics $g$ (which is also sometimes called the Yamabe invariant). I feel like my lack of visualisation of conformal changes of metric is stopping me from answering what might be a very basic question:
Given a Riemannian manifold $(M,g)$ of dimension $\geq 3$, with $Y(g)>0$, does every metric in the conformal class of $g$ (i.e. those metrics $e^{2f}g$ for smooth functions $f$) have everywhere positive scalar curvature? $(*)$
In other words (by the Yamabe problem), if $(M,g)$ admits a conformal metric of constant positive scalar curvature, does every conformal metric have positive scalar curvature? As a potential counterexample, could one find a metric conformal to the standard metric on the sphere which has non-positive scalar curvature somewhere?
If the answer to $(*)$ is no, could there exist a metric in the conformal class of $g$ with everywhere negative scalar curvature?
If $\tilde g = e^{2 \varphi}g $ are conformally related metrics, the corresponding scalar curvatures are related by the transformation law $$\tilde S = e^{-2\varphi}\left(S + 2(n-1)\Delta_g\varphi - (n-2)(n-1)|\nabla\varphi|_g^2\right).\tag 1 $$
Thus the fact that $g$ has (e.g.) $S=1$ places no pointwise constraint on $\tilde S$: constructing a smooth function $\varphi$ to have a local maximum of $1$ at $p$ but very quickly fall nearby, we can make $\tilde S(p)$ negative with arbitrarily large magnitude. Thus the answer to $(*)$ is no.
The answer to your followup question, however, might be more satisfying: If $S\ge 0$ everywhere and $\varphi$ attains its minimum at some point $p$ (e.g. if $M$ is compact), then at $p$ we have (again from $(1)$) $$\tilde S = e^{-2 \varphi}(S + 2(n-1)\Delta_g \varphi) \ge e^{-2 \varphi} S \ge 0,$$ so at least in the compact case, a metric with (everywhere) positive scalar curvature cannot be conformally related to one with (everywhere) negative scalar curvature.