There is no hope of showing that $w N. E(\mathcal C) \to w N. \mathcal C \times w N. \mathcal C$ is a homotopy equivalence without assuming that the cofibrations of $\mathcal C$ are splittable. To see that, take a simple example and look at the direct sum Grothendieck groups which appear as $\pi_1$ of these K-theory spaces. Let $\mathcal C$ be the category of finitely generated abelian groups with the non-split short exact sequence $M \in E(\mathcal C)$ given by $0 \to \mathbb Z \to \mathbb Z \to \mathbb Z/2 \to 0$, and consider the class $[M]$ in $K_0^\oplus \mathcal C$. From an equation $[M]=[M']$, where $M'$ is a splittable short exact sequence, we can deduce that $M$ and $M'$ are stably isomorphic, i.e., there is another short exact sequence $N$ with $M \oplus N \cong M' \oplus N$. The functor $S_2 \mathcal C \to \mathcal C$ that computes the kernel of the tensor product with $\mathbb Z/2$ of the first map in the short exact sequence is an additive functor that sends $M$ to $\mathbb Z/2$, $M'$ to zero, and $N$ to a finite abelian 2-group, yielding an isomorphism between two finite abelian groups of different order, giving a contradiction.

There is no hope of showing that $w N. E(\mathcal C) \to w N. \mathcal C \times w N. \mathcal C$ is a homotopy equivalence without assuming that the cofibrations of $\mathcal C$ are splittable. To see that, take a simple example and look at the direct sum Grothendieck groups which appear as $\pi_1$ of these K-theory spaces. Let $\mathcal C$ be the category of finitely generated abelian groups with the non-split short exact sequence $M \in E(\mathcal C)$ given by $0 \to \mathbb Z \to \mathbb Z \to \mathbb Z/2 \to 0$, and consider the class $[M]$ in $K_0^\oplus \mathcal C$. From an equation $[M]=[M']$, where $M'$ is a splittable short exact sequence, we can deduce that $M$ and $M'$ are stably isomorphic, i.e., there is another short exact sequence $N$ with $M \oplus N \cong M' \oplus N$. The functor $E( \mathcal C ) \to \mathcal C$ that computes the kernel of the tensor product with $\mathbb Z/2$ of the first map in the short exact sequence is an additive functor that sends $M$ to $\mathbb Z/2$, $M'$ to zero, and $N$ to a finite abelian 2-group, yielding an isomorphism between two finite abelian groups of different order, giving a contradiction.

There is no hope of showing that $w N. E(\mathcal C) \to w N. \mathcal C \times w N. \mathcal C$ is a homotopy equivalence without assuming that the cofibrations of $\mathcal C$ are splittable. To see that, take a simple example and look at the direct sum Grothendieck groups which appear as $\pi_1$ of these K-theory spaces. Let $\mathcal C$ be the category of finitely generated abelian groups with the non-split short exact sequence $M \in S_2 \mathcal C$ given by $0 \to \mathbb Z \to \mathbb Z \to \mathbb Z/2 \to 0$, and consider the class $[M]$ in $K_0^\oplus \mathcal C$. From an equation $[M]=[M']$, where $M'$ is a splittable short exact sequence, we can deduce that $M$ and $M'$ are stably isomorphic, i.e., there is another short exact sequence $N$ with $M \oplus N \cong M' \oplus N$. The functor $S_2 \mathcal C \to \mathcal C$ that computes the kernel of the tensor product with $\mathbb Z/2$ of the first map in the short exact sequence is an additive functor that sends $M$ to $\mathbb Z/2$, $M'$ to zero, and $N$ to a finite abelian 2-group, yielding an isomorphism between two finite abelian groups of different order, giving a contradiction.

There is no hope of showing that $w N. E(\mathcal C) \to w N. \mathcal C \times w N. \mathcal C$ is a homotopy equivalence without assuming that the cofibrations of $\mathcal C$ are splittable. To see that, take a simple example and look at the direct sum Grothendieck groups which appear as $\pi_1$ of these K-theory spaces. Let $\mathcal C$ be the category of finitely generated abelian groups with the non-split short exact sequence $M \in E(\mathcal C)$ given by $0 \to \mathbb Z \to \mathbb Z \to \mathbb Z/2 \to 0$, and consider the class $[M]$ in $K_0^\oplus \mathcal C$. From an equation $[M]=[M']$, where $M'$ is a splittable short exact sequence, we can deduce that $M$ and $M'$ are stably isomorphic, i.e., there is another short exact sequence $N$ with $M \oplus N \cong M' \oplus N$. The functor $S_2 \mathcal C \to \mathcal C$ that computes the kernel of the tensor product with $\mathbb Z/2$ of the first map in the short exact sequence is an additive functor that sends $M$ to $\mathbb Z/2$, $M'$ to zero, and $N$ to a finite abelian 2-group, yielding an isomorphism between two finite abelian groups of different order, giving a contradiction.