Let me offer another viewpoint.  By a Feynman-Kac formula, a solution to $$
\partial_t m(t, y) = -b \partial_y m(t,y) + \frac{1}{2} \sigma^2 \partial_y^2 m(t,y) \;, ~~ m(0,y) = m_0(y)\;, ~~ m(t,0) = m(t,1) = 0\;,
$$ admits the following stochastic representation
$$
m(t, y) = \mathbb{E}_y \{ 1_{\{\tilde  T \ge t\}} m_0(Z_t) \} 
$$
where $\mathbb{E}_y$ is an expected value over the process which satisfies
$$
Z_t = y - b t + \sigma W_t
$$
and $\tilde  T$ is the first exit time of $Z$ from $(0,1)$. Alternatively, one can write $m(t,y)$ as
$$
m(t, y) = \mathbb{E}_y  \{  m_0(Z_t) \mid \tilde  T \ge t \} \mathbb{P}_y(\tilde T \ge t)  
$$
Unfortunately, due to the truncation of $X$ by $T$, there does not seem to be a relation between this $m(t,y)$ and the transition density of $Y$.