I understand your question as follows.
Is there a constant $U$ that if $G$ is a finite group with the center $C(G)$ of order two and centralizers $C(a)$ of index two for every $a \in G \setminus C(G)$, then the order of $G$ is bounded above by $U$?
The answer to the above question is negative; and extraspecial $2$-groups provide a counterexample. Since such a group has the center of order two and the order is not bounded, all we have to prove is that the indices of centralizers of non-central elements equal two.
Let $G$ be an extraspecial $2$-group with center $C(G) = \langle t \rangle$ and $a \in G \setminus C(G)$. Because $G' = C(G)$, we have $[a, g] \in \langle t \rangle$ for every element $g \in G$. In other words, a map
$$ \phi \colon G \to G', \quad g \mapsto [a, g]$$
is surjective. Hence $G = \phi^{-1}(1) \sqcup \phi^{-1}(t)$. As $\phi^{-1}(1) = C(a)$, we need to investigate what $\phi^{-1}(t)$ is. Let $b \in G \setminus C(a)$. Note that $[a, b] = t$. We claim that $\phi^{-1}(t) = bC(a)$. Suppose $g \in bC(a)$ and write $g = bc$ for some $c \in C(a)$. Then
$$[a, g] = [a, bc] = c^{-1}[a, b]c = c^{-1}tc = t.$$
Hence $g \in \phi^{-1}(t)$.
Next, suppose $g \in \phi^{-1}(t)$. Then
$$
\begin{align*}
[a, b^{-1}g]
&= a^{-1}g^{-1}bab^{-1}g
= a^{-1}g^{-1}bab^{-1}(a^{-1}a)g\\
&= a^{-1}g^{-1}[b^{-1}, a^{-1}]ag
= a^{-1}g^{-1}ag[b^{-1}, a^{-1}]\\
&= [a, g][b^{-1}, a^{-1}]
= t[b^{-1}, a^{-1}]
= [a, b][b^{-1}, a^{-1}]\\
&= a^{-1}b^{-1}ab[b^{-1}, a^{-1}]
= a^{-1}b^{-1}[b^{-1}, a^{-1}]ab
= a^{-1}b^{-1}(bab^{-1}a^{-1})ab \\
&= 1.
\end{align*}
$$
Therefore, $b^{-1}g \in \phi^{-1}(1) = C(a)$.
Now we proved that $G = C(a) \sqcup bC(a)$, so that $C(a)$ has index two.
