FALSE: the pumping lemma characterizes the regular languages

The language $G$ satisfies the pumping conclusion with pumping length $4$. If $s=w{w}^{R}v\in G$ with $\mathrm{\mid }w\mathrm{\mid }\ge 2$, write $w=au$ with $u\mathrm{\ne }\epsilon$ and choose $x=\epsilon$, $y=a$, $z=u{u}^{R}av$; then $xz=u{u}^{R}av\in G$, the original word is $xyz\in G$, and for every $i\ge 2$ the word $x{y}^{i}z=a^{i}u{u}^{R}av$ lies in $G$ by using the one-letter witness word $a$ and the nonempty suffix $a^{i-2}u{u}^{R}av$.

For each $n\ge 1$, let $p_n:=(01{)}^n$ and $x_n:=(10{)}^{n}1$. Then $p_n{x}_n=(01{)}^n(10{)}^{n}1$ lies in $G$ by taking witness word $(01{)}^n$ and suffix $1$. If $m>n$ and $p_m{x}_n$ were in $G$, write $p_m{x}_n=r{r}^{R}v$ with $r\mathrm{\ne }\epsilon$ and $v\in \{0,1{\}}^{+}$. The word $p_m{x}_n=(01{)}^m(10{)}^{n}1$ has exactly one doubled adjacent pair, namely the middle 11 at the junction between $(01{)}^m$ and $(10{)}^{n}1$. Any prefix of the form $r{r}^R$ has a doubled middle pair at the boundary between $r$ and $r^R$, so that boundary must be this unique 11. Hence $\mathrm{\mid }r\mathrm{\mid }=2m$, which makes the palindromic prefix $r{r}^R$ have length $4m$. But $\mathrm{\mid }{p}_m{x}_n\mathrm{\mid }=2m+2n+1<4m+1$ because $n<m$, so no such factorization with nonempty $v$ exists. Therefore $p_m{x}_n\mathrm{\notin }G$, and the prefixes $p_n$ are pairwise distinguishable.

If instead $\mathrm{\mid }w\mathrm{\mid }=1$, then $s=aav$ with $\mathrm{\mid }v\mathrm{\mid }\ge 2$ because $\mathrm{\mid }s\mathrm{\mid }\ge 4$; writing $v=bt$ and choosing $x=aa$, $y=b$, $z=t$ pumps entirely inside the nonempty suffix $v$, so every $x{y}^{i}z$ stays in $G$. Thus every sufficiently long word in $G$ has a legal pumping decomposition.

Step 1.2 and [L2] show that $G$ is not regular, while steps 1.1 and 2.1 show that $G$ still has the regular pumping property. Therefore the pumping property does not characterize regularity, and the statement is false.