Does the de Rham complex induce a functorial soft resolution of the category of cochain complexes of sheaves of vector spaces on a smooth manifold?

I apologize in advance if this is pretty straightforward; I'm a differential geometer and physicist by training so my homological algebra and homotopy theory are a bit weak.

Question: Let $M$ be a smooth manifold, $\Omega^\bullet_M$ the de Rham complex of sheaves, and $\mathbb{R}_M$ the constant sheaf corresponding to $\mathbb{R}$. Does the tensor product of chain complexes of sheaves $(-) \otimes \Omega_M$ with the de Rham complex give a functorial soft resolution of the category $\mathsf{Chain}(\mathbb{R}_M \mathsf{Mod})$ of chain complexes of sheaves of vector spaces on $M$?

If this is true, does anyone happen to know of a reference for this?

If this is not true in general, is it true for some appropriate subcategory, such as bounded chain complexes, or bounded below chain complexes? I suspect that in fact this will be the case for bounded below complexes of sheaves.

Here's my (partial) reasoning for why I suspect the proposition in the question should be true:

I remember reading somewhere (probably in Kashiwara & Schapira, but can't find the result at the moment) that if a sheaf $\mathcal{F}$ is c-soft and flat, then $\mathcal{F} \otimes \mathcal{A}$ is c-soft for any other sheaf $\mathcal{A}$. I assume this should still be true if we replace the words "sheaf" by "chain complex of sheaves", so that both $\mathcal{A}^\bullet$ and $\mathcal{F}^\bullet$ are complexes of sheaves, and $\mathcal{F}^\bullet$ is degree-wise flat and c-soft.

Assuming this is true, this should give the result I want, since sheaves of vector spaces are always flat, and so the de Rham complex is degree-wise c-soft and flat. In detail, we have a quasi-isomorphism $\mathbb{R}_M \xrightarrow{\sim} \Omega_M$ by the usual de Rham resolution. Since tensoring sheaves of vector spaces is exact, then for any complex of sheaves of vector spaces $\mathcal{A}^\bullet \in \mathsf{Chain}(\mathbb{R}_M \mathsf{Mod})$ tensoring with $\mathcal{A}^\bullet$ should preserve the quasi-isomorphism, and hence give us a quasi-isomorphism $\mathcal{A}^\bullet \cong \mathcal{A}^\bullet \otimes \mathbb{R}_M \xrightarrow{\sim} \Omega_M^\bullet \otimes \mathcal{A}^\bullet$.

Is this reasoning correct?

For some motivation about why it would be interesting to know if this is true, in Riehl's Categorical Homotopy Theory text, she shows that given a functorial right deformation $R: \mathsf{M} \to \mathsf{M}$, $r: 1 \overset{\sim}{\Rightarrow} R$ for a functor $F: \mathsf{M} \to \mathsf{N}$ between homotopical categories, we can compute a point-set right derived functor $\mathbb{R}F = F \circ R: \mathsf{M} \to \mathsf{N}$. This point-set derived functor depends on the choice of deformation, but descends to the usual (deformation-independent) total derived functor $\mathbf{R}F: \mathrm{Ho}(\mathsf{M}) \to \mathrm{Ho}(\mathsf{N})$ between the homotopy categories.

In my case we would have $\mathsf{M} = \mathsf{Chain}(\mathbb{R}_M \mathsf{Mod})$ and we would want to compute point-set derived functors for $f_!$ and $\Gamma(M,-)$ using the de Rham complex.