We formally define a uniformly elliptic operator on a $C^1$ bounded domain $\Omega\subset\mathbb R^d$ as follows. Let $a^{ij},b^i,c\in L^\infty(\Omega)$, let $(a^{ij})$ be a symmetric uniformly positive-definite matrix, and define $$L = -\sum_{i,j}a^{ij}\partial_i\partial_j + \sum_i b^i\partial_i + c.$$ We define $H^1(\Omega)$ as usual to be the functions with $f,\nabla f\in L^2$ weakly, $H_0^1$ to be those functions in $H^1$ with zero trace, and $H^{-1}(\Omega):=H^1_0(\Omega)'$.
In Evans' PDE book, at the end of section 6.1, we seem to implicitly be assuming that $L$ maps $H^1$ to $H^{-1}$. However, I don't see why we can do this without assuming that the $a^{ij}$ are in $W^{1,\infty}$, i.e. they are Lipschitz continuous. Otherwise, do we not run the risk of multiplying an order $1$ distribution, or even just a dirac delta mass, by an $L^\infty$ function, and getting something undefined, or in general unbounded?
Evans defines (see (2) in Section 6.1) $$ Lu = -\sum_{i,j} \partial_j( a^{ij} \partial_i u) + \sum_i b^i \partial_i u + c \, u. $$ This is to be understood as an element in $H^{-1}$ via $$ \langle L u , v \rangle = \int_\Omega \sum_{i,j} a^{ij} \, \partial_i u \, \partial_j v + \sum_i b^i \, \partial_i u \, v + c \, u \, v \,\mathrm{d}x $$ for $u \in H^1$ and $v \in H_0^1$.
Your definition of $L$ is similar to the non-divergence form of $L$ given in (3) in Section 6.1, see also the remark following (2) and (3).
(I use the second edition of Evans' book).