It follows from Theorem 2 of the following paper and the geometrization theorem that $Y^3\times S^1$ admits a metric of positive scalar curvature iff $Y$ does.

*Schoen, Richard; Yau, Shing-Tung*, **On the structure of manifolds with positive scalar curvature**, Manuscr. Math. 28, 159-183 (1979). ZBL0423.53032.

On p. 9 of the paper, they define $C_3'$ to be the class of closed three dimensional manifolds which do not
admit any non-zero degree map to a compact three dimensional manifold
$M$ such that $\pi_2(M) = 0$ and $M$ contains a two-sided incompressible
surface with genus $\geq 1$ (i.e. a Haken 3-manifold).
It follows from the geometrization theorem and virtual Haken theorem that $Y$ admits a metric of positive scalar curvature iff every finite-sheeted cover $Y'\to Y$ has $Y'\in C_3'$.

Let $C_4'$ be the class of closed 4-manifolds $M$ such that every codimension one homology
class of $M$ can be represented, up to some non-zero integer, by a map
from a manifold of class $C'_3$ (a homology class $\alpha\in H_k(M)$ is represented by an oriented $k$-manifold $N$ if there is a map $f:N\to M$ such that $f_\ast([N])=\alpha$, where $[N]\in H_k(N)$ is the fundamental class determined by the orientation of $N$).

Then Theorem 2 implies that if $M^4$ admits a metric of positive scalar curvature,
then $M$ is of class $C'_4$.

Now, consider the 4-manifold $Y^3\times S^1$ and let $Y'\to Y$ be a finite-sheeted
cover. By Theorem 2, there exists an $N\subset Y'\times S^1$ such that $[N]=[Y'\times \ast]\in H_3(Y'\times S^1)$ and $N$ is of class $C_3'$. Moreover, there is a non-zero degree map $N\to Y'$ (by composing the embedding with the projection $Y'\times S^1\to Y'$), hence $Y'\in C_3'$ (the class $C_3'$ is clearly closed under taking non-zero degree maps). Since every finite-sheeted cover of $Y$ is in $C_3'$, we conclude that $Y$ admits a metric of positive scalar curvature.