The dimesion is indeed one.

One can show that the element
\begin{align*}
\alpha_1 &= 1,\\
\alpha_n &= (1-n)\int_0^1 \frac{t^{n-1}}{\log(1-t)} dt\quad\text{for } n = 2,3,\dots
\end{align*}
lies in $E$.
So let $x\in E$ with $x_1=0$. We need to show that $x=0$. 
Let $y_k=\frac {k+1}{k}x_{k+1}$.
Then $y\in\ell^\infty$ and $Tx=x$ as well as $x_1=0$ lead to
\begin{align*}
0&=\sum_{k=1}^\infty\frac{y_{k}}{k(k+1)},\\
\frac m{m+1}y_m&=\sum_{k=m+1}^\infty\frac{y_k}{(k-m)(k-m+1)},\quad m\ge 1.
\end{align*}
We have to show $y=0$.
For $m\in{\mathbb N}_0$ let $z^{(m)}\in\ell^1$ be defined by
\begin{align*}
z^{(0)}_k&=\frac1{k(k+1)},\\
z^{(m)}_k&=\begin{cases}0&k<m,\\
\frac m{m+1}&k=m,\\
\frac{-1}{(k-m)(k-m+1)}&k>m.
\end{cases}
\end{align*}
Now $\ell^\infty$ is the dual space of $\ell^1$.
Let $\langle .,.\rangle$ denote the duality pairing.
The formulas above say that
$$
\langle{z^{(k)},y}=\rangle0\qquad\text{for }k=0,1,\dots
$$
Therefore the theorem follows if we can show that the sequence $z^{(0)},z^{(1)},\dots$ spans  a dense subspace of $\ell^1$.
Let $Z$ denote the span of $z^{(0)},z^{(1)},\dots$. We will now show that the first canonical basis element $e_1=(1,0,\dots)$ lies in the closure of $Z$.
We have
$$
\begin{array}{ccccccccc}
z^{(1)}&=&\frac12,&\frac{-1}2,&\frac{-1}{6},&\frac{-1}{12},&\frac{-1}{20},&\frac{-1}{30},&\dots\\
&\\
z^{(2)}&=&0,&\frac23,&\frac{-1}2,&\frac{-1}{6},&\frac{-1}{12},&\frac{-1}{20},&\dots\\
&\\
z^{(3)}&=&0,&0,&\frac34,&\frac{-1}2,&\frac{-1}{6},&\frac{-1}{12},&\dots\\
&\\
z^{(4)}&=&0,&0,&0,&\frac45,&\frac{-1}2,&\frac{-1}{6},&\dots\\
&\\
z^{(5)}&=&0,&0,&0,&0,&\frac56,&\frac{-1}2,&\dots\\
\end{array}
$$
Let $a^{(1)},a^{(2)},\dots\in\ell^1$ be defined  by $a^{(1)}=z^{(1)}$ and for $m\ge 2$ inductively by
\begin{align*}
a^{(m)}_k=\begin{cases}
\frac12&k=1,\\
0&2\le k\le m,\\
a^{(m-1)}_k+a_m^{(m-1)}\frac{m+1}m\frac1{(k-m)(k+1-m)}& k\ge m+1.
\end{cases}
\end{align*}
Then $a^{(m)}=a^{(m-1)}-a_m^{(m-1)}\frac{m+1}mz^{(m)}$, so that $a^{(m)}\in Z$.
Iterating the recursive relation, we get
$$
a^{(m)}=a^{(1)}-a_2^{(1)}\frac32 z^{(2)}-\dots-a_m^{(m-1)}\frac{m+1}m z^{(m)}.
$$
Setting $b_j=a_{j+1}^{(j)}$ this becomes
$$
a^{(m)}=z^{(1)}-\sum_{j=1}^{m-1}b_j\frac{j+2}{j+1}z^{(j+1)}.
$$
If we can show that the sequence $a^{(m)}$ converges in $\ell^1$, it follows that it converges to $\frac12e_1$, where $e_1=(1,0,0,\dots)$ is the first standard basis vector, so that we have
$$
e_1\in\overline Z.
$$
As the $\ell^1$-norms of ths $z^{(j)}$ are globally bounded, in order to show convergence of $(a^{(m)})$, it suffices to show $\sum_{j=1}^\infty|b_j|<\infty$.
For this we read the recursive relation for $k>m$ as
$$
a^{(m)}_k=\frac{-1}{k(k-1)}+\sum_{j=1}^{m-1}b_j\frac{j+2}{j+1}\frac1{(k-j-1)(k-j)}.
$$
In particular, for $k=m+1$ we get
$$
b_m=\frac{-1}{m(m+1)}+\sum_{j=1}^{m-1}b_j\frac{j+2}{j+1}\frac1{(m-j)(m+1-j)}.
$$
It follows $b_j<0$ for every $j$ and with $c_j=|b+j|$ we have
\begin{align*}
c_m&=\frac1{m(m+1)}+\sum_{j=1}^{m-1}c_j\frac{j+2}{j+1}\frac1{(m-j)(m+1-j)}\\
&=\frac1{m(m+1)}+\sum_{j=1}^{m-1}c_j\frac{j+2}{j+1}\frac1{(m+1-j)(m+2-j)}\underbrace{\frac{m+2-j}{m-j}}_{\le\frac32}\\
&\le\frac1{m(m+1)}+\frac32\sum_{j=1}^{m-1}c_j\frac{j+2}{j+1}\frac1{(m+1-j)(m+2-j)}.
\end{align*}
Using $\frac1{m(m+1)}=\frac1m-\frac1{m+1}$ and collapsing the ensuing telescope sums we get
\begin{align*}
\sum_{m=1}^Nc_m&\le\sum_{m=1}^N\frac1{m(m+1)}
+\frac32\sum_{j=1}^{N-1}c_j\frac{j+2}{j+1}\sum_{m=j+1}^N\frac1{(m+1-j)(m+2-j)}\\
&=\left(1-\frac1{N+1}\right)+\frac32\sum_{j=1}^{N-1}c_j\frac{j+2}{j+1}
\left(\frac12-\frac1{N+2-j}\right)\\
&\le 1+\frac34\sum_{j=1}^{N-1}c_j\left(1+\frac{1}{j+1}\right).
\end{align*}
Subtracting $\frac34\sum_{j=1}^{N-1}c_j$ and multiplying by 4 we get
\begin{align*}
\sum_{m=1}^Nc_m&\le \sum_{m=1}^Nc_m+3c_N\\
&\le 4+3\sum_{j=1}^{N-1}\frac{c_j}{j+1}\\
&=4+\frac32c_1+c_2+\frac34c_3+\dots+\frac3{N}c_{N-1}\\
&\le4+\frac34c_1+\frac14c_2+\frac34\sum_{j=1}^{N}c_j.
\end{align*}
Subtracting $\frac34\sum_{j=1}^{N-1}c_j$ and multiplying by 4 we get
\begin{align*}
\sum_{m=1}^Nc_m&\le 16+3c_1+c_2.
\end{align*}
Since this holds for every $N$, we conclude $\sum_{m=1}^\infty c_m<\infty$.
So we have shown $e_1\in\overline Z$. The same argument mutatis mutandis shows $ej\in\overline Z$ for all $j$.
The claim follows.