The construction of adding $p$ as an additional element to the vector $\mathbf u$ only hides it in the vector $\mathbf v$, without actually eliminating it from the Euler equation. This can be easily done with a vector $\mathbf e=(0,0,\ldots 0,0,1)$, and a diagonal matrix $\Delta={\rm diag}\,(1,1,,\ldots 1,1,0)$ so that $\Delta\cdot\mathbf v=(\mathbf u,0)$ and $\mathbf e\cdot\mathbf v=p$, but such a construction seems rather pointless:
$$\partial_t (\Delta\cdot\mathbf v)+(\Delta\cdot\mathbf v)D(\Delta\cdot\mathbf v)=D(\mathbf e\cdot\mathbf v).$$
To really eliminate the pressure from the Euler equation you can use the Leray projection, which maps a vector field to its zero-divergence part: If $\mathbf v=\mathbf w+\nabla\phi$ with $\text{div}\,\mathbf w=0$, then the Leray projection is $\mathbb P(\mathbf v)=\mathbf w$. Formally, this construction can be written as
$$\mathbb P(\mathbf u) = \mathbf u - \nabla \Delta^{-1} (\nabla \cdot \mathbf u).$$
The Euler equation then reads
$$\frac{\partial}{\partial t}\mathbf u+\mathbb P[\mathbf u\cdot\nabla\mathbf u]=0.$$
More explicitly, this can be written as (see for example page 6 of these notes):
$$\frac{\partial}{\partial t}\mathbf u(\mathbf x,t)+\mathbf u(\mathbf x,t)\cdot\nabla\mathbf u(\mathbf x,t)=\frac{1}{C_d}\int_{\mathbb R^d}\frac{\mathbf x- \mathbf y}{|\mathbf x-\mathbf y|^d}\sum_{i,j}\frac{\partial u_i(\mathbf y,t)}{\partial y_j}\frac{\partial u_j(\mathbf y,t)}{\partial y_i}\,d\mathbf y,$$
with $C_d$ the surface area of the $d$-dimensional unit sphere.