Here is a partial answer, I will prove the following:

If $M\times \mathbb{S}^1$ equipped with locally $\mathrm{CAT}(0)$ length metric, then there is isometric $\mathbb{S}^1$-action on $M\times \mathbb{S}^1$ with parallel geodesic orbits.

Denote by $N$ the universal cover of $M\times \mathbb{S}^1$;
it is a $\mathrm{CAT}(0)$ length metric.
Let $\gamma$ be a shortest circle in $M\times \mathbb{S}^1$ that is homotopic to $p\times \mathbb{S}^1$.
Denote by $\tilde\gamma_\alpha$ lifts of $\gamma$ in $N$; each $\tilde\gamma_\alpha$ is a line in $N$.
Denote by $b_\alpha$ the Busemann function associated to $\gamma_\alpha$.

Note that projection of $\gamma_\alpha$ to $\gamma_\beta$ is isometric for any $\alpha$ and $\beta$.
It follows that $b_\alpha-b_\beta$ is constant for any $\alpha$ and $\beta$.
Therefore, the gradient $\nabla b_\alpha$ is independent of $\alpha$ and invariant with respect to the action of deck transformations on $N$.

It follows that $M\times \mathbb{S}^1$ admits a vector field $v$ such that $\nabla b_\alpha$ is a lift of $v$ for any $\alpha$. (Gradient flow on singular spaces is discussed in our book.)

Note that the flow $\Phi^t$ along $v$ is distance-noncontracting.
Since it is defined on a closed manifold $M\times \mathbb{S}^1$, $\Phi^t$ is isometry for any $t$.
It follows that $b_\alpha$ is affine (it is convex and concave at the same time).
By the line strip theorem, $N$ splits isometrically as $L\times \mathbb{R}$.
Passing back to $M\times \mathbb{S}^1$, we get the statement.