= Solution
For fixed $\mathbf R$ and $t$, define $h(\mathbf x,s)=n_2(\mathbf a,\mathbf b,t)$ with $\mathbf a=\mathbf x+(s-1)\mathbf R$ and $\mathbf b=\mathbf x+s\mathbf R$. The <chain rule> gives
$$
\partial_sh=\mathbf R\cdot(\nabla_{\mathbf a}+\nabla_{\mathbf b})n_2,\qquad \nabla_{\mathbf x}h=(\nabla_{\mathbf a}+\nabla_{\mathbf b})n_2,
$$
so $\boxed{\partial_sh=\mathbf R\cdot\nabla_{\mathbf x}h}$. The derivative with respect to $\mathbf x$ here translates \i[both] arguments of the <two-particle number density>; it is not a derivative of its first argument alone.
Take the spatial divergence of the proposed <interaction pressure tensor>. Its <pair potential> factor is independent of $\mathbf x$, so the preceding identity gives
$$
\begin{aligned}
\partial_{x_j}P^\phi_{ij}
&=-\frac12\int d^3R\,\frac{R_i\phi'(R)}R\int_0^1\partial_sh\,ds\\
&=-\frac12\int d^3R\,\frac{R_i\phi'(R)}R\left[n_2(\mathbf x,\mathbf x+\mathbf R)-n_2(\mathbf x-\mathbf R,\mathbf x)\right].
\end{aligned}
$$
Exchange symmetry of the <two-particle distribution function> implies $n_2(\mathbf x-\mathbf R,\mathbf x)=n_2(\mathbf x,\mathbf x-\mathbf R)$. In the second term, substitute $\mathbf R\mapsto-\mathbf R$. Since $R_i\phi'(R)/R$ is odd, both terms become equal:
$$
\partial_{x_j}P^\phi_{ij}=-\int d^3R\,\frac{R_i\phi'(R)}R\,n_2(\mathbf x,\mathbf x+\mathbf R).
$$
On the other hand, at fixed $\mathbf x_\star=\mathbf x+\mathbf R$, the first-argument derivative of the <pair potential> is $\nabla_{\mathbf x}\phi=-\mathbf R\phi'(R)/R$. Thus the interaction <force density> in <1/b> is $-\partial_{x_j}P^\phi_{ij}$. Moving it to the left yields local <conservation of momentum>:
$$
\boxed{\partial_t(\rho\mathbf u)+\nabla\cdot(\rho\mathbf u\mathbf u+\mathbf P+\mathbf P^\phi)=0.}
$$
The integral over $s$ distributes each pair's <momentum flux> along the straight segment joining the particles. The factor $1/2$ removes the double counting of ordered pairs. Central <forces> also make $P^\phi_{ij}=P^\phi_{ji}$.
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