This can be a little confusing at first. Let's call the basis vectors of $B$: $e_1=(1,1)$ and $e_2=(-1,0)$. First of all, $e_1$ and $e_2$ are given to us in terms of the standard basis (that is, $e_1=(1,1)$ where $(1,1)$ is a vector in the standard basis). In the basis $B$, the vector $(1,0)$ really means $(1,0)=1*e_1+0*e_2=e_1$ and so $(1,0)$ in $B$ corresponds to the vector $(1,1)$ in the standard basis. Similarly, $(0,1)$ in $B$ corresponds to the vector $(-1,0)$ in the standard basis.
This gives us a change of basis matrix:
$$
C=\begin{pmatrix}
1&-1\\1&0
\end{pmatrix}
$$
which takes us from basis $B$ to the standard basis (i.e. check that $C*(1,0)$ gives you $(1,1)$). Naturally, $C^{-1}$ will take you from the standard basis to $B$. So given a vector $v$ in the standard basis we have:
$$
Tv=C[T]_BC^{-1}v
$$
Note that $C^{-1}v$ represents $v$ in the basis $B$. We know the matrix for $T$ in $B$ so $[T]_BC^{-1}v$ gives us $Tv$ in the basis $B$. Finally we use $C$ to get back to the standard basis. So:
$$
[T]_E=C[T]_BC^{-1}= \begin{pmatrix} 1&0 \\ 0&0 \end{pmatrix}
$$