Yes.
First note that Galois connections are far more common than field theorists would have you believe: any binary relation gives rise to one and the two fixed sets a closed under any appropriate algebraic structure.
Write
$P\perp Q$ for "every member of $P$ is orthogonal (in the sense of factorisation systems) to every member of $Q$";
$P;Q$ for the set of all composites of members of $P$ followed by members of $Q$;
$L'=L\cap E_0\cap M_1$ and $R'=R\cap E_0\cap M_1$ (to avoid any ambiguity in this argument); and
$E=E_1;L'$ and $M=R';M_0$.
Then
$E_1\perp M_1\supset R'$,
$E_1\perp M_1\supset M_0$,
$L'\subset L\perp R\supset R'$ and
$L'\subset E_0\perp M_0$.
Since orthogonality respects composition, it follows that $E\perp M$, whilst by construction $E;M$ is the entire hom-class of the category.
Therefore $(E,M)$ is a factorisation system.
The things that I have assumed are all proved in my book but I don't have a copy to hand to look up the references.
In the question as stated, it is given that $M_0\subset M_1$, but there is no order-relationship between these and $R$. This is why it was necessary to introduce $R'$. The construction gives a way of handling expressions in the lattice of factorisation systems on a category, so it would be an interesting exercise in lattice theory to find out whether this is modular or even distributive.