It is not true. Take $\mathfrak{sl}_3$, with simple reflections $s,t$, such that $s$ is singular. Put $x=e$, $w= st$; both are in $W^\Sigma$.

I claim that $Ext_\mathcal{O}^i(M(\mu),L(st \cdot \mu)) = 0$ for all $i \geq 0$.

Denote by $\lambda$ a regular, integral antidominant weight. By [2, Corollary 1.3.3.] we have

$$ \dim Ext_\mathcal{O}^i(M(\mu),L(st \cdot \mu)) = \begin{cases} \dim Hom(M(\lambda),L(st \cdot \lambda))=0  & \colon i=0 \\
\dim Ext_\mathcal{O}^i(M(\lambda),L(st \cdot \lambda)) - \dim Ext_\mathcal{O}^{i-1}(M(s \cdot \lambda),L(st \cdot \lambda)) &:i>0 \end{cases}$$.

We know that $P_{e,st} = P_{s,st} = 1$. From KL polynomial in terms of extension group in the regular case [1, Theorem 8.11.], it follows that the above always cancels. (For $i=2$ we have $1-1=0$.)

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[1] Humphreys: Representations of semisimple Lie algebras in the BGG category O. Graduate Studies in Mathematics, 94. American Mathematical Society, Providence, RI, 2008.

[2] Irving: Singular blocks of the category O, Math Z (1990) 204: 209.

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EDIT: General explanation: A singular block of $\mathcal{O}$ is Koszul-dual to a regular block of a parabolic category $\mathcal{O}^{\mathfrak{p}}$, where $\mathfrak{p}$ is given by the same singularity set. By this duality, the dimensions of Ext-group correspond to the multiplicities of simples inside generalized Verma modules. But strange things may happen with simples inside generalized Vermas, in particular, some simples that are "expected" to appear by the Bruhat order, may not appear. This is because generalized Verma is a quotient of the usual Verma, where you kill all submodules that are not in $\mathcal{O}^{\mathfrak{p}}$. But a lot of composition factors that actually exist in $\mathcal{O}^{\mathfrak{p}}$ are killed in the quotient.