I believe the following is a counterexample:
Let $N=1$, $B_1(0)=(-1,1)$, $u(x)=|x|$, then $|u^\prime(x)| = 1$, $u^{\prime \prime}(x) = 2\delta_0$,
and
$u^\prime(x) = 2H(x)-1$,
where $H$ is the Heaviside function, $H \in L^\infty \cap W^{1,2}_{loc}(-1,1)$.W^{1,2}_{loc}((-1,1)\setminus {0})$.
Thus $u$ solves $(\frac{1}{2} u^\prime)^\prime = H^\prime$,
which is the PDE in one dimension, with $H$ in the right space, $u$ is smooth away from the origin and $u \in W^{1,\infty}(-1,1)$, but $u^{\prime\prime}$ is only a measure and so not in $L^p(-1,1)$ for any $p>1$.
In particular, $u$ is not $C^1$, $u$ is not smooth at the origin, and $u^{\prime\prime} \notin L^p(-1,1)$.
More generally, the computation extends into more dimensions, as far as I can tell.
