Can we find a non central element of order 2 in a specific 2-group? Let $G$ be a non-abelian group of order $2^5$ and center $Z(G)$ is non cyclic. Can we always find an element $x\not\in Z(G)$ of order $2$ if for any pair of elements $a$ and $b$ of $Z(G)$ of order $2$, the factor groups $G/\langle a\rangle$ and $G/\langle b\rangle$ are isomorphic?
Note: I do not find any contradiction of above statement so I wish to prove it.

This is not complete answer; but a partial information, which in addition to Holt's comment may simplify your job. (I will try to write complete proof as I get some directions on it)
Suppose $G$ is a group of order $2^5$ satisfying conditions in your question. We show that $Z(G)=C_2\times C_2$. 
For this, by hypothesis, $|Z(G)|=2^2$ or $2^3$. If $|Z(G)|=2^3$, then $G/Z(G)=C_2\times C_2$. Hence there exists $x,y\in G$ be such that $G=\langle x,y,Z(G)\rangle$. Then $G'=\langle [x,y]\rangle$ and order of this commutator subgroup must be $2$ (fact: if $G/Z(G)$ is abelian then $G/Z(G)$ and $G'$ have same exponent.) Note that $G/Z(G)$ is abelian implies $G'\subseteq Z(G)$.
Since $Z(G)$ is non-cyclic, it should contain a subgroup of order $2$ other than $G'$; call it $N$. Then $G/N$ is non-abelian, whereas $G/G'$ is abelian. Thus, we are getting two subgroups of order $2$ in $Z(G)$ with non-isomorphic quotient.
Thus $|Z(G)|$ must be $4$ and it should be then $C_2\times C_2$ (it is non-cyclic).
