Let $L_{1,}L_{2}$ be regular languages and define $L:=\{uw \mid \exists v\in\Sigma^{*}:uv\in L_{1},vw\in L_{2}\}$.
I wish to prove that $L$ is regular using only closure properties (such as $L_{1},L_{2}$ is regular then so is $L_{1}\cap L_{2},L1\cup L_{2},L_{1}^{*}$ etc.).
My thoughts: I tried to define a homomorphism $h:\Sigma\cup\Sigma'\to\Sigma^{*}$
$\forall\sigma\in\Sigma:h(\sigma)=\sigma$ $\forall\sigma'\in\Sigma':h(\sigma')=\epsilon$
Then I tried to look at $h^{-1}(L_{1}),h^{-1}(L_{2})$, I thought about looking at their intersection with the languages $\Sigma^{*}(\Sigma')^{*},(\Sigma')^{*}\Sigma^{*}$ etc to give the words in $h^{-1}(L_{1}),h^{-1}(L_{2})$ the form of $ww'$s.t $w\in L_{1}$ or $w'w$ s.t $w\in L_{2}$.
I am having difficulty to continue, I thought about trying to get to something like $L_{3}(\Sigma')^{*}L_{4}$ where $L_{3}$ is all the prefix of words in $L_{1}$ and $L_{4}$ is all the endings of words in $L_{2}$ and then intersect this with $\Sigma^{*}L_{5}'\Sigma^{*}$ where $L_{5}$ is some language to enforce that the words in $(\Sigma')^{*}$in the expression$L_{3}(\Sigma')^{*}L_{4}$ will be such that if I remove all the tags (using a homomorphism) I would get $L$.
How can I show $L$ is regular using only closure properties ?