The linear transformation $T$ takes vectors in $\mathbb{R}^3$ to vectors in $\mathbb{R}^3$, so $T(e_2)$ needs to be a vector in $\mathbb{R}^3$ and not an element of $\Lambda^2 (\mathbb{R}^3)$. Specifically,
$$
T(e_2) = \begin{pmatrix}
1 & 0 & 1\\
1 & 1 & 0\\
0 & 1 & 1 \end{pmatrix}\begin{pmatrix} 0\\ 1 \\ 0 \end{pmatrix}
=\begin{pmatrix}
0 \\ 1 \\ 1 \end{pmatrix} = e_2 + e_3.
$$
Similarly, we get that $T(e_1) = e_1 + e_2$, so we must have that
$$
T(e_1) \wedge T(e_2) = (e_1 + e_2) \wedge (e_2 + e_3).
$$
Edit: To address your follows-up question, recall that to find the matrix of the linear operator $\Lambda^2(T)$ with respect to the basis $\{ e_1 \wedge e_2, e_1 \wedge e_3, e_2 \wedge e_3 \}$, it suffices to find the image of the basis vectors under $\Lambda^2(T)$. The first such computation is done above, giving that
$$
\Lambda^2(T) (e_1 \wedge e_2) = e_1 \wedge e_2 + e_1 \wedge e_3 + e_2 \wedge e_3.
$$
Let's compute the image of the next basic vector:
\begin{align}
\Lambda^2(T)(e_1 \wedge e_3)
&= T(e_1) \wedge T(e_3)\\
&= (e_1 + e_2) \wedge (e_1 + e_3)\\
&= e_1 \wedge e_1 + e_1 \wedge e_3 + e_2 \wedge e_1 + e_2 \wedge e_3\\
&= -e_1 \wedge e_2 + e_1 \wedge e_3 + e_2 \wedge e_3.
\end{align}
Similarly, we find that $\Lambda^2(T)(e_2 \wedge e_3) = -e_1 \wedge e_2 - e_2 \wedge e_3 + e_2 \wedge e_3$. Recall that the first column of the matrix $\Lambda^2(T)$ will be the image of the first basis vector $e_1 \wedge e_2$, i.e. the vector $(1,1,1)$; the second column of the matrix will be the image of the second basis vector $e_1 \wedge e_3$, i.e. the vector $(-1,1,1)$; the third column of the matrix is the image of the third basis vector $e_2 \wedge e_3$, i.e. $(-1,-1,1)$. Therefore, the matrix of $\Lambda^2(T)$ with respect to the basis $\{ e_1 \wedge e_2, e_1 \wedge e_3, e_2 \wedge e_3 \}$ is given by
$$
\Lambda^2(T) = \begin{pmatrix}
1 & -1 & -1 \\
1 & 1 & -1 \\
1 & 1 & 1 \end{pmatrix}.
$$
