Here are the details behind my original interpretation of Deane Yang's comment:

By Schauder estimates, $\vert\vert \nabla u\vert\vert_{L^{\infty}} \leq C\vert\vert u\vert\vert_{L^{\infty}}$, 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 $Lu = \sum_{i,j=1}^n \partial_j(a^{ij}(x)\partial_i u) + \sum_{i=1}^nb^i(x)\partial_i u + c(x)u$. 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 $0$. (EDIT: The dimension should be at least $2$ for this to work)

Now $L^2$ elliptic regularity implies that $u$ is smooth on the whole ball and we are done.