I am adding the lines below after having read the comments.
Sorry for having been so\dots elliptic in my answer.I should have added the following details: first we choose $\varphi \geq 0$, so that the critical points $u_{k}$ are nonnegative. Then according to Gidas-Ni-Nirenberg's result $u_{k}$ has a spherical symmetry, that is $u_{k}$ is radial. Next, since $$\Vert \nabla u_{k}\Vert^2 + \Vert u_{k}\Vert^2 = \Vert u_{k}\Vert_{p}^{p} \lesssim \Vert \nabla u_{k}\Vert^{\theta p}\Vert u_{k}\Vert^{(1-\theta)p},$$(using Gagliardo-Nirenberg inequality, for some $0<\theta<1$) then one deduces that there is $R_{0} >0$ such that for all $k \geq1$ we have $\Vert u_{k}\Vert_{p}^{p} = \Vert \nabla u_{k}\Vert^2 + \Vert u_{k}\Vert^2 \geq R_{0}^2$. Using the fact that the imbedding $H^1_{\rm rad}({\Bbb R}^n) \subset L^{p}({\Bbb R}^n)$ is compact, one infers that $u_{k}\to {\overline u}$ strongly in $L^{p}$, and weakly in $H^1$. In particular $\Vert {\overline u}\Vert_{p}^{p}\geq R_{0}^2$, and thus ${\overline u}\not\equiv 0$. Using the equation satisfied by $u_{k}$ and the strong convergence in $L^{p}$, one checks easily that ${\overline u}$ is solution to (E), and hence $$\Vert \nabla {\overline u}\Vert^2 + \Vert {\overline u}\Vert^2 = \Vert {\overline u}\Vert_{p}^{p},$$yielding also that the convergence in $H^1$ is strong.

