Here are the details behind Deane Yang's comment:
By Schauder estimates, $\vert\vert \nabla u\vert\vert_{L^{\infty}} \leq \vert\vert u\vert\vert_{L^{\infty}}$, so we also know that the gradient of $u$ is uniformly bounded. Next, we claim that $u$ is a weak solution of $Lu = 0$ on $B_1(0)$. We have $Lu = \sum_{i,j=1}^n a^{ij}(x)\partial_{ij}^ 2 u + \sum_{i=1}^nb^i(x)\partial_i u + c(x)u$. To show that $u$ is a weak solution on $B_1(0)$, we need to show that for every $\varphi \in C_c^{\infty}(B_1(0))$,
$\int_{B_1(0)} \sum_{i,j=1}^n a^{ij}(x)\partial_{i}u\partial_j\varphi + \sum_{i=1}^nb^i(x)(\partial_i u) \varphi+ c(x)u\varphi = 0$
Since all of the relevant terms are uniformly bounded, the left hand side is equal to
$\lim_{\epsilon\to 0}\int_{B_1(0)\setminus B_{\epsilon}(0)} \sum_{i,j=1}^n a^{ij}(x)\partial_{i}u\partial_j\varphi + \sum_{i=1}^nb^i(x)(\partial_i u) \varphi+ c(x)u\varphi$
After an integration by parts, this gives
$\lim_{\epsilon\to 0}\int_{B_1(0)\setminus B_{\epsilon}(0)}(Lu)\varphi + \int_{\partial B_{\epsilon}(0)}\sum_{i,j=1}^n a^{ij}(x)(\partial_{i}u)\varphi \nu_j = $
$\lim_{\epsilon\to 0}\int_{\partial B_{\epsilon}(0)}\sum_{i,j=1}^n a^{ij}(x)(\partial_{i}u)\varphi \nu_j$
Since everything involved is uniformly bounded, this goes to $0$.
Now $L^2$ elliptic regularity implies that $u$ is smooth on the whole ball and we are done.