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 $$ m(t, y) = \mathbb{E}_y \{ 1_{\{T \ge t\}} m_0(Z_t) \} $$ where $\mathbb{E}_y$ is an expected value over the process which satisfies $$ Z_t = y - b t + \sigma W_t $$ and $T$ is the first exit time of $Z$ from $(0,1)$. Alternatively, one can write $m(t,y)$ as $$ m(t, y) = \mathbb{E}_y \{ m_0(Z_t) \mid T \ge t \} \mathbb{P}_y(T \ge t) $$ Due to the truncation of $X$, there does not seem to be a simple relation between this $m(t,y)$ and the transition density of $Y$.
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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 $$ \partial_t m(t, y) = -b \partial_y m(t,y) + \frac{1}{2} \sigma^2 \partial_y^2 m(t,y) \;, ~~ m(0,y) = m_0(y)\;. $$ 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 $$ \int_{\mathbb{R}} m_0(\xi) \left( - \partial_t p(t,y,\xi) + b \partial_{\xi} p(t,y,\xi) + \frac{1}{2} \sigma^2 \partial_{\xi}^2 p(t,y,\xi) \right) d\xi = 0 $$ If this holds for all $m_0 \in C_0^2(\mathbb{R})$, then one gets the Fokker-Planck equation $$ \partial_t p(t,y,\xi) = b \partial_{\xi} p(t,y,\xi) + \frac{1}{2} \sigma^2 \partial_{\xi}^2 p(t,y,\xi) $$ If I'm not mistaken, this procedure can't be repeated in the bounded case under consideration.