Does the following operation makes the language regular?
Interesting question. As Orlp pointed out, $L$ does not contain 01. So it cannot be $(0+1)^*$. 

In fact $L$ is not regular. Consider $M=L\cap L(0^+110^+1)$.

Claim. $M=\{0^i110^i1\mid i\gt0\}$.
Proof. Since $(0^i1)1(0^i1)\in M$ for $i&gt;0$, we have the "$\supseteq$". Now suppose $wxw=00^i110^j1$, $w\in\{0,1\}^+$, $x\in\{ 0,1\}^*$, $i,j\ge0$. then $w$ must starts with 0 and ends with 1. Since there only three 1's in $wxw$, $w=00^i1$ and $x=1$. So $wxw=00^i1100^i1$, $i\ge0$. We have the "$\subseteq$".

There is a counting requirement for words in $M$. So, intuitively, $M$ cannot be regular. I will let you prove it rigorously using, for example, the pumping lemma for regular language.

Since $M$ is not regular and $0^+110^+1$ is regular, $L$ cannot be regular.