The answer to this question is no. 

Indeed it holds that $$P(Y_t = 1) = P\big( \max\limits_{s\le t} X_s \ge 1\big) > 0.$$
So the density has an atom in $1$ and similarly in $0$.

Let us derive some heuristics. You are right in implicitly saying that the Brownian Motion (plus a possible drift and variance) with absorbing boundary has as a generator the operator $\mathcal{G}$ with Dirichlet boundary conditions. And from this indeed it follows that the probability density solves the adjoint equation, related to $\mathcal{G}^*$ in the interior of the domain. But since *you are testing only against functions with Dirichlet boundary conditions* you cannot deduce anything about $p(t,0)$ or $p(t,1).$

On the other hand, one can ask whether the only bad things are the atoms. I.e., whether the absolutely continuous part w.r.t. to the Lebesgue measure solves the forward Kolmogorov equation with Dirichlet boundary.

In this case, if my calculations are correct, the answer is positive, at least for $\sigma =1$ and $b = 0$. Indeed we would be interested in the limit $$\lim_{\epsilon \to 0} m(t, 1-\epsilon)$$ Forgetting the lower boundary, which does not play a role in this, and replacing $1$ with $R$ for clarity, we can compute:
$$
m(t, R-\epsilon) \le \lim_{h \to 0 } \frac{1}{h}\mathbb{P}(X_t \in ( R{-}\epsilon{-}h, R{-}\epsilon), M_t < R)
$$
where $M_t$ is the running maximum of $X_t$.
Now the joint distribution of $X_t$ and $M_t$ is well known through the reflection principle. If my calculations are correct the last term can be computed as
$$
\lim_{h \to 0 } \frac{1}{h}\mathbb{P}(X_t \in ( R{-}\epsilon{-}h, R{-}\epsilon), M_t < R)= \varphi_t(R-\epsilon) - \varphi_t(R+\epsilon)
$$
where $\varphi_t$ is the density of a $N(0,t)$ random variable. Taking the limit over $\epsilon$ gives the result. In the case of drift and diffusion one has to use a more general formula for the distribution density of the process and it's running maximum (unless one comes up with an intelligent trick).  Such a formula can be found for example [here][1] but I was to lazy to do the calculations.

EDIT: in the spirit of finding some heuristics, I believe that one can think of the second case as follows: the probability of being near to the boundary without the maximum ever hitting the boundary vanishes, I would guess due to the fluctuations of the B.M.. It would be like asking for the probability of assuming the maximum exactly at time $t$.

  [1]: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.486.3934&rep=rep1&type=pdf