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 $Y$.

**Add**

In this addendum, I briefly explain how $m(t,y)$ is related to the transition density of $Z$ in the *unbounded* case where, as @kenneth states, $$
m(t, y) = \mathbb{E}_y m_0(Z_t) 
$$ satisfies the forward Kolmogorov equation $$
\partial_t m(t, y) = -b \partial_y m(t,y) + \frac{1}{2} \sigma^2 \partial_y^2 m(t,y) \;, ~~ m(0,y) = m_0(y)\;.
$$  To obtain the corresponding Fokker-Planck equation for the transition density $p(t,y,\xi)$ of the process $Z$, one "differentiates under the integral sign" and integrates by parts to get \begin{align*}
\partial_t \mathbb{E}_y m_0(Z_t)  &= \int_{\mathbb{R}}  m_0(\xi) \partial_t p(t,y,\xi) d\xi \\
&= \int_{\mathbb{R}}  \left( - b  m_0'(\xi) + \frac{1}{2} \sigma^2  m_0''(\xi) \right) p(t,y,\xi) d\xi \\
&= \int_{\mathbb{R}} m_0(\xi) \left(  b \partial_{\xi} p(t,y,\xi) + \frac{1}{2} \sigma^2 \partial_{\xi}^2 p(t,y,\xi) \right) d\xi
\end{align*}
 Hence, we get $$
\int_{\mathbb{R}} m_0(\xi) \left( - \partial_t p(t,y,\xi) +  b \partial_{\xi} p(t,y,\xi) + \frac{1}{2} \sigma^2 \partial_{\xi}^2 p(t,y,\xi) \right) d\xi = 0
$$ for all $m_0 \in C_0^2(\mathbb{R})$.  Thus, $$
\partial_t p(t,y,\xi) =  b \partial_{\xi} p(t,y,\xi) + \frac{1}{2} \sigma^2 \partial_{\xi}^2 p(t,y,\xi)
$$  If I'm not mistaken, this procedure can't be repeated in the bounded case under consideration.