Showing $T$ is continuous.

$T$ is not continuous. For this it suffices to consider $p_n(t)=1-(2t-1)^{2n}$. Then $\|p_n\|=1$, but $\|Tp_n\|= Tp_n(1)=2\sum_{j=1}^{n}\frac{1}{2j-1}$. Since
$$ \|Tp_n\|\ge S_n:=\sum_{j=1}^{n}\frac 1j \quad\text{and}\quad\lim_{n\to\infty}S_n=+\infty $$ we can take $q_n=p_n/S_n$ and then $\displaystyle\lim_{n\to\infty}\|q_n\|=0$ but $\|Tq_n\|\ge 1$ for all $n$.

The key observation to understand what's going on here conceptually is that the operator $T$ is given by $$Tp(x)=p(0)+\int_{0}^x\frac{p(t)-p(0)}{t}\, dt.$$ This makes it easy to see that in fact $T$ is unbounded, essentially since the integral $\int_0^1\frac{1}{t}\, dt$ diverges. For instance, for any $\epsilon>0$, we can find (by the Weierstrass approximation theorem) a polynomial $p$ such that $\|p\|\leq 1$, $p(0)=0$, and $p(t)\geq 1-\epsilon$ for all $t\in [\epsilon, 1]$. We then have $$Tp(1)-Tp(\epsilon)=\int_{\epsilon}^1\frac{p(t)}{t}\,dt\geq\int_{\epsilon}^1\frac{1-\epsilon}{t}\,dt=(1-\epsilon)\log(1/\epsilon).$$ This goes to infinity as $\epsilon$ goes to $0$, and so the norms of the $Tp$ also go to infinity. So there is no bound on $\|Tp\|$ for polynomials $p$ such that $\|p\|=1$, and so $T$ is unbounded.

$\newcommand{\norm}[1]{\lVert#1\rVert}$Note: This is not a complete answer but might be a way to approach the answer.

By Stone-Weierstrass, $\mathcal P$'s uniform closure is $\mathrm C[0,1]$. (The space of continuous functions on $[0,1]$)

Define, for each $n \in \mathbb{W}$, $$ T_n\left( \sum_{j=0}^m a_j x^j \right) = a_0 +\sum_{j=1}^n \frac{a_j}{j} x^j $$ where trailing $a_j$'s (Those with indices greater than $m$ but less or equal to $n$) are set to $0$. Then each $T_n$ is bounded via triangle inequality.

If you can somehow show that $T_n$ is a Cauchy sequence in $\operatorname{CHom}[\mathcal P \to \mathrm C[0,1]]$ (the space of continuous linear maps), a theorem in functional analysis would imply that the pointwise limit of $\langle T_n \rangle$ is $T$ and $T$ is bounded in $\norm\cdot$.