Center not is quotient-local powering-invariant in solvable group

Statement
It is possible to have a solvable group $$G$$ with center $$Z(G)$$ such that $$Z(G)$$ is not a quotient-local powering-invariant subgroup of $$G$$. Explicitly, let $$\varphi:G \to G/Z(G)$$ be the quotient map. Then, there exists a prime number $$p$$ and an element $$g \in G$$ such that $$g$$ has a unique $$p^{th}$$ root in $$G$$, but $$\varphi(g)$$ has multiple $$p^{th}$$ roots in $$G/Z(G)$$.

Proof
Define:

$$G := \langle x,y,z \mid y^2 = x^2z, [x,z] = [y,z] = 1 \rangle$$

We can understand the structure of $$G$$ using the following normal series:

$$1 \le \langle z \rangle \le \langle x^2, z \rangle \le \langle x^2, xy, z \rangle \le \langle x,y,z \rangle$$

The successive quotients are $$\mathbb{Z}, \mathbb{Z}, \mathbb{Z}, \mathbb{Z}/2\mathbb{Z}$$. More details below:


 * $$\langle z \rangle$$ is the center and the quotient group $$G/\langle z \rangle$$ is isomorphic to the amalgamated free product $$\mathbb{Z} *_{2\mathbb{Z}} \mathbb{Z}$$, with the two pieces generated by the images of $$x$$ and $$y$$ and the amalgamated part being the image of $$x^2$$, which coincides with the image of $$y^2$$.
 * $$\langle x^2, z\rangle$$ is the second center and the quotient group $$G/\langle x^2,z \rangle$$ is isomorphic to the free product $$\mathbb{Z}/2\mathbb{Z} * \mathbb{Z}/2\mathbb{Z}$$, which in turn is isomorphic to the infinite dihedral group (where the images of $$x$$ and $$y$$ are both reflections whose product gives a generator for the cyclic maximal subgroup).
 * $$\langle x^2, xy, z \rangle/\langle x^2, z\rangle$$ is the cyclic maximal subgroup inside $$G/\langle x^2,z \rangle$$.

Consider now the element $$g = x^2$$. We have the following:


 * $$G$$ is solvable: This is obvious from the normal series for $$G$$ where all the quotients are abelian.
 * $$g$$ has a unique square root, namely $$x$$, in $$G$$: This requires some work to show rigorously, and can be demonstrated using a polycyclic presentation with the elements $$x,u,z$$ where $$u = xy$$. The idea is to compute the general expression for the square of an arbitrary element that is of the form $$x^{m_1}u^{m_2}z^{m_3}$$ and deduce that, for the square to equal $$g$$, we must have $$m_1 = 1, m_2 = m_3 = 0$$.
 * Under the quotient map by the center, the image of $$g$$ does not have a unique square root: The center is $$\langle z \rangle$$, and the quotient group is $$\mathbb{Z} *_{2\mathbb{Z}} \mathbb{Z}$$. As noted above, the image of $$g = x^2$$ coincides with the image of $$y^2$$ in the quotient, but the images of $$x$$ and $$y$$ do not coincide. Thus, the image of $$g$$ modulo the center does not have a unique square root in the quotient group by the center.