You tagged your question with "c-star-algebras" and "von-neumann-algebras" but you obviously seem to be mostly interested in the W-crossed product, rather than its C-version. Nevertheless let me give a counter example in the C*-case, hoping that it may be modified to fit your interest.
Let $X$ be a set with two points, none of which has measure zero, so that $L^∞(X)$ is isomorphic to $ℂ^2$. Let $G:=ℤ$ act on $X$ trivially, so that $$ L^∞(X)⋊G = ℂ^2⋊ℤ = ℂ^2⊗C^*(ℤ) = ℂ^2⊗C(S^1) = C(S^1) ⊕ C(S^1) = C(S^1\sqcup S^1), $$ where the square cup stands for disjoint union.
For each $n$ in $Z$, let $u_n$ be the corresponding implementing unitary in $L^∞(X)⋊G$. Note that, if $u_n$ is viewed as an element of $C(S^1\sqcup S^1)$, one has that $$ u_n(z) = z^n, $$ for every $z$ in either copy of $S^1$.
Let $x_1$ and $x_2$ be the two versions of "1" in the two copies of $S^1$, and consider the subalgebra $B$ of $C(S^1\sqcup S^1)$ formed by all functions $f$ such that $f(x_1)=f(x_2)$.
Since $u_n(x_1)=u_n(x_2)$, for all $n$, we have that $u_n$ lies in $B$, so that $$C^*(ℤ)⊆B,$$ which I assume is the appropriate C*-interpretation of your condition that $L(G)⊆B$.
Nevertheless $B$ is not a crossed product by $ℤ$, because its spectrum is a figure eight, lacking a free action of $S^1$ to play the role of the dual action.