For $c=0$ this result is knows as reflection principle (see e.g. René Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Chapter 6) and follows from the Markov property and symmetry of Brownian motion. However, for $c>0$ the proof is more involved since we have to get rid of the drift term.
Since by definition
$$[H_a \leq t] = \left[ \sup_{s \leq t} X_s \geq a \right] \tag{1}$$
determining the distribution of $H_a$ is equivalent to finding the distribution of $\sup_{s \leq t} X_s$. In order to find the distribution of the latter, we need two ingredients: Girsanov's theorem and the joint distribution $(X_t,\sup_{s \leq t} X_s)$ for a Brownian motion $(X_t)_{t \ge 0}$.
Girsanov theorem: Let $c \in \mathbb{R}$ and $(B_t)_{t \geq 0}$ be a Brownian motion on a probability space $(\Omega,\mathcal{A},\mathbb{P})$. Then $$X_t := B_t+ct, \qquad t \leq T,$$ is a Brownian motion with respect to the probability measure $$d\mathbb{Q} := d\mathbb{Q}_T := \exp \left( -c B_T - \frac{c^2}{2} T \right) d\mathbb{P}.$$
For a proof see e.g. René Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Chapter 18.
Joint distribution of $(X_t, \sup_{s \leq t} X_s)$: Let $(X_t)_{t \geq 0}$ be a Brownian motion on a probability space $(\Omega,\mathcal{A},\mathbb{Q})$. Then the joint distribution $(X_t,\sup_{s \leq t} X_s)$ equals $$\mathbb{Q} \left[ X_t \in dx, \sup_{s \leq t} X_s \in dy \right] = \frac{2 (2y-x)}{\sqrt{2\pi t^3}} \exp \left(- \frac{(2y-x)^2}{2t} \right) 1_{[-\infty,y]}(x) \, dx \, dy. \tag{2}$$
For a proof see e.g. René Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Exercise 6.8 (there are full solutions available on the web).
So let's finally put it all together: It follows from $(1)$ and the definition of the probability measure $\mathbb{Q}_T$ that
$$\begin{align*} \mathbb{P}[H_a \leq T] &= \mathbb{P} \left[ \sup_{s \leq T} X_s \geq a \right] = \int 1_{[a,\infty)} \left( \sup_{s \leq T} X_s \right) \, d\mathbb{P} \\ &= \int 1_{[a,\infty)}\left( \sup_{s \leq T} X_s \right) \exp \left( c B_T + \frac{c^2}{2} T \right) \, d\mathbb{Q}_T \\ &= \int 1_{[a,\infty)}\left( \sup_{s \leq T} X_s \right)\exp\left(c X_T- \frac{c^2}{2} T \right) \, d\mathbb{Q}_T. \end{align*}$$
By Girsanov's theorem, $(X_t)_{t \leq T}$ is a Brownian motion with respect to $\mathbb{Q}_T$ and therefore $(2)$ gives
$$\begin{align*} \mathbb{P}[H_a \leq T] &= \exp\left(- \frac{c^2}{2} T \right) \int_{y \geq a} \int_{x \leq y} e^{cx} \frac{2 (2y-x)}{\sqrt{2\pi T^3}} \exp \left(- \frac{(2y-x)^2}{2T} \right) \, dx \, dy. \end{align*}$$
It remains to calculate the integral expression. First of all, by Fubini's theorem,
$$\begin{align*} \mathbb{P}[H_a \leq T] &= \frac{1}{\sqrt{2\pi T}} \exp \left(- \frac{c^2}{2} T \right) \left(\int_{x \geq a} e^{cx} I_1(x) \, dx + \int_{x \leq a} e^{-cx} I_2(x) \, dx \right) \\ &:=J_1+J_2 \tag{3} \end{align*}$$
where
$$\begin{align*} I_1(x):= \int_{y \geq x} \frac{2(2y-x)}{T} \exp \left(- \frac{(2y-x)^2}{2T} \right) \, dy &= \left[ - \exp \left(- \frac{(2y-x)^2}{2T} \right) \right]_{y=x}^{\infty} \\ &= \exp \left(- \frac{x^2}{2T} \right) \\ I_2(x) := \int_{y \geq a} \frac{2(2y-x)}{T} \exp \left(- \frac{(2y-x)^2}{2T} \right) \, dy &=\exp \left(- \frac{(2a-x)^2}{2T} \right). \end{align*}$$
Hence,
$$\begin{align*} J_1 &= \frac{1}{\sqrt{2\pi T}} \exp \left(- \frac{c^2}{2} T \right) \int_{x \geq a} e^{cx} I_1(x) \, dx \\ &= \frac{1}{\sqrt{2\pi T}} \int_{x \geq a} \exp \left(- \frac{(x-cT)^2}{2T} \right) \, dx \\ &= \frac{1}{\sqrt{\pi}} \int_{z \geq \frac{a-cT}{\sqrt{2T}}} \exp(-z^2) \, dz \tag{4} \end{align*}$$
and
$$\begin{align*} J_2 &= \frac{1}{\sqrt{2\pi T}} \exp \left(- \frac{c^2}{2} T \right) \int_{x \leq a} e^{cx} I_2(x) \, dx \\ &\stackrel{u:=2a-x}{=} \frac{1}{\sqrt{2\pi T}} \exp \left(- \frac{c^2}{2} T \right) \int_{u \geq a} e^{c(2a-u)} \exp\left(-\frac{u^2}{2T} \right) \, du \\ &=\ldots = \frac{e^{2ac}}{\sqrt{\pi}} \int_{z \geq \frac{a+CT}{\sqrt{2T}}} \exp(-z^2) \, dz. \tag{5} \end{align*}$$
Now if we differentiate $(3)$ with respect to $T$, using $(4)$ and $(5)$, we get
$$\begin{align*} \frac{d}{dT} \mathbb{P}(H_a \leq T) &= \frac{-1}{\sqrt{\pi}} \exp \left( - \frac{(a-cT)^2}{2T} \right) \left( \frac{-c}{\sqrt{2T}} - \frac{a-cT}{2 \sqrt{2} T^{3/2}} \right) \\ &\quad + \frac{-1}{\sqrt{\pi}} \underbrace{e^{2ac} \exp \left( - \frac{(a+cT)^2}{2T} \right)}_{\exp(-(a-cT)^2/2T)} \left( \frac{c}{\sqrt{2T}} - \frac{a+cT}{2 \sqrt{2} T^{3/2}} \right) \\ &= \frac{a}{\sqrt{2\pi T^3}} \exp \left(- \frac{(a-cT)^2}{2T} \right). \end{align*}$$
Let $W$ be standard
Brownian motion and $Y$ geometric
Brownian motion, i.e., for $t\geq 0$
$$
Y_{t}=y_{0}\exp (\alpha t+\sigma W_{t}),
$$
where $y_{0}\in \mathbb{R}_{++}$, $\alpha \in \mathbb{R}$, and $\sigma \in
\mathbb{R}_{++}$.
The Handbook of Brownian Motion by Borodin and Salminen says that for any $y\geq y_{0}$ and $r>0$
$$
\Bbb{E}[e^{-r\tau _{y}}]=\left(\frac{y_{0}}{y}\right)^{\kappa},
$$
where $\tau_y$ is the time of the first transition to $y$ and
$$
\kappa =\left(\sqrt{\alpha ^{2}+2r\sigma ^{2}}-\alpha \right)\sigma ^{-2}
$$
is a strictly positive constant.
Which I think means you are close.
Best Answer
I have now found all mistakes, so here is the final answer.
We use separation of variables to find a solution we define $u(x,t) = f(x)g(t)$ then we get the two equations: \begin{align*} \begin{cases} D\partial_x^2 f - \mu \partial_x f + \lambda f = 0 \\ \partial_t g = - \lambda g \end{cases} \end{align*} where we have introduced the constant $-\lambda$ and where $D=σ^2/2$. We start by solving the time dependent equation, this equation has the general solution: \begin{align*} g(t) = g(t_0) \exp\big(-\lambda(t-t_0) \big) \ . \end{align*} For the spatial dependent solution $f$, we put $f(x) = \exp(rx)$ then we get: \begin{align*} D r^2 \exp(rx) - \mu r \exp(rx) + \lambda \exp(rx) = 0 \ . \end{align*} We divide by $\exp(rx)$ and solve the corresponding second order polynomial: \begin{align*} r_1 &= \frac{\mu + \sqrt{\mu^2 - 4 D \lambda}}{2 D} \ , \\ r_2 &= \frac{\mu - \sqrt{\mu^2 - 4 D \lambda}}{2 D} \ . \end{align*} Now if $\mu^2 - 4 D \lambda>0$ and $r_1 \neq r_2$ is not equal then we have the solution: \begin{align*} f(x) = A \exp(r_1 x) + B \exp(r_2 x) \ . \end{align*} If $\mu^2 - 4 D \lambda=0$ then $r_1 = r_2$ and we then have the solution: \begin{align*} f(x) = A \exp(r x) + B x\exp(r x) \ . \end{align*} If $\mu^2 - 4 D \lambda<0$ then the roots are complex, thus we have $r_1 = \frac{\mu + i \sqrt{4 D \lambda - \mu^2}}{2D}$, $r_2 = \frac{\mu - i \sqrt{4 D \lambda - \mu^2}}{2D}$ and $\beta = \frac{\sqrt{4 D \lambda - \mu^2}}{2D}$. The solution is: \begin{align*} f(x) = \exp\left(\frac{\mu x}{2D}\right) \left(A \sin(\beta x) + B \cos(\beta x) \right) \ . \end{align*} Using the boundary condition only the last expression for $f$ will give us a non-trivial solution. Thus we have: \begin{align*} f(a) &= \exp\left(\frac{\mu a}{2D}\right) \left(A \sin(\beta a) + B \cos(\beta a) \right) = 0 \ \Rightarrow \ A = - B \frac{\cos(\beta a)}{ \sin(\beta a)} \ . \end{align*} Using the other boundary point we get: \begin{align*} f(b) &= \exp\left(\frac{\mu b}{2D}\right) \left( - B \frac{\cos(\beta a)}{ \sin(\beta a)} \sin(\beta b) + B \cos(\beta b) \right) \ . \end{align*} Form this we conclude that: \begin{align*} B \left ( \cos(\beta b) - \frac{\cos(\beta a)}{ \sin(\beta a)} \sin(\beta b) \right) = 0 \ . \end{align*} If $B$ is zero we get a trivial solution, thus $B$ is not zero and then we conclude that $\cos(\beta b) - \frac{\cos(\beta a)}{ \sin(\beta a)} \sin(\beta b) = 0$. By multiplication with $-\sin(\beta a)$ and using the following formula: \begin{align*} \sin(\beta b)\cos(\beta a) - \cos(\beta b) \sin(\beta a) = \sin(\beta ( b - a) ) \ , \end{align*} we see that: \begin{align*} \sin(\beta ( b - a) ) = 0 \ . \end{align*} This is true for $\beta ( b - a) = n \pi$ where $n \in \mathbb{Z}$. By the formula $\beta = \frac{\sqrt{4 D \lambda - \mu^2}}{2D}$ we see that there is not just one possible constant $\lambda$, but one for every $n \in \mathbb{Z}$: \begin{align*} \lambda_n = D \left( \left( \frac{n \pi}{b-a} \right)^2 + \left( \frac{\mu}{2D} \right)^2 \right) \ , \ \ n \in \mathbb{Z} \ . \end{align*} Where we have introduced a subscript $n$ to the possible $\lambda$ constants to separate them. Now by the linearity of the equation we get a Fourier series as a solution: \begin{align*} u(x,t) = \exp\left(\frac{\mu x}{2D}\right) \sum_{n \in \mathbb{Z}} \left(A_n \sin\left(\frac{n \pi x}{b-a} \right) + B_n \cos\left(\frac{n \pi x}{b-a} \right) \right) \exp\big(-\lambda_n(t-t_0) \big) \ . \end{align*} Where $g(t_0)$ has been left out since it can be absorbed into the constants $A_n$ and $B_n$. Now because of the odd and even properties of sine and cosine we see that: \begin{align*} A_n \sin\left(\frac{n \pi x}{b-a} \right) + A_{-n} \sin\left(\frac{-n \pi x}{b-a} \right) &= (A_n - A_{-n}) \sin\left(\frac{n \pi x}{b-a} \right) \ , \ n \in \mathbb{N} \ , \\ B_n \cos\left(\frac{n \pi x}{b-a} \right) + B_{-n} \cos\left(\frac{-n \pi x}{b-a} \right) &= (B_n + B_{-n}) \cos\left(\frac{n \pi x}{b-a} \right) \ , \ n \in \mathbb{N} \ . \end{align*} Because of these formulas and the squared $n^2$ in the expression of $\lambda_n$ we can limit our sum to $n \in \mathbb{N}_0$ thus: \begin{align*} u(x,t) = \exp\left(\frac{\mu x}{2D}\right) \sum_{n \in \mathbb{N}_0} \left(A_n \sin\left(\frac{n \pi x}{b-a} \right) + B_n \cos\left(\frac{n \pi x}{b-a} \right) \right) \exp\big(-\lambda_n(t-t_0) \big) \ . \end{align*} One can show the following for sine and cosine: \begin{align*} \int_{a}^b \cos\left(\frac{n \pi x}{b-a} \right) \sin\left(\frac{m \pi x}{b-a} \right) dx &= 0 \ , \ \forall n,m \ . \\ \int_a^b \cos\left(\frac{n \pi x}{b-a} \right) \cos\left(\frac{n \pi x}{b-a} \right) dx &= 0 \ , \ n \neq m \ . \\ \int_a^b \sin\left(\frac{n \pi x}{b-a} \right) \sin\left(\frac{n \pi x}{b-a} \right) dx &= 0 \ , \ n \neq m \ . \\ \end{align*} For the two last expressions, in the case $m=n$, we have: \begin{align*} \int_a^b \cos^2\left(\frac{n \pi x}{b-a} \right) dx &= \frac{b-a}{2} \ , \\ \int_a^b \sin^2\left(\frac{n \pi x}{b-a} \right) dx &= \frac{b-a}{2} \ . \end{align*} From these formulas one can deduce that: \begin{align*} B_n = \frac{2}{b-a} \exp\big(\lambda_n(t-t_0) \big) \int_a^b u(x,t) \cos\left(\frac{n \pi x}{b-a} \right) \exp\left(\frac{-\mu x}{2D}\right) dx \ , \\ A_n = \frac{2}{b-a} \exp\big(\lambda_n(t-t_0) \big) \int_a^b u(x,t) \sin\left(\frac{n \pi x}{b-a} \right) \exp\left(\frac{-\mu x}{2D}\right) dx \ . \end{align*} Since this has to be true for all $t \geq t_0$ we can put $t=t_0$ then we get: \begin{align*} B_n = \frac{2}{b-a} \int_a^b \delta(x-x_0) \cos\left(\frac{n \pi x}{b-a} \right) \exp\left(\frac{-\mu x}{2D}\right) dx = \frac{2}{b-a} \cos\left(\frac{n \pi x_0}{b-a} \right) \exp\left(\frac{-\mu x_0}{2D}\right) \ , \\ A_n = \frac{2}{b-a} \int_a^b \delta(x-x_0) \sin\left(\frac{n \pi x}{b-a} \right) \exp\left(\frac{-\mu x}{2D}\right) dx = \frac{2}{b-a} \sin\left(\frac{n \pi x_0}{b-a} \right) \exp\left(\frac{-\mu x_0}{2D}\right) \ . \end{align*} Inserting this back into our solution we get: \begin{align*} u(x,t) = \frac{2}{b-a} \exp\left(\frac{\mu(x-x_0)}{2D}\right) \sum_{n \in \mathbb{N}_0} \cos\left(\frac{n \pi (x-x_0)}{b-a} \right) \exp\big(-\lambda_n(t-t_0) \big) \end{align*} We now define the survival probability as: \begin{align*} S(t) = \int_a^b u(x,t) dx \ . \end{align*} We see that: \begin{align*} S(t) &= \frac{2}{b-a} \sum_{n \in \mathbb{N}_0} \exp\big(-\lambda_n(t-t_0) \big) \int_a^b \exp\left( \frac{\mu (x-x_0)}{2D} \right) \cos\left( \frac{n \pi (x-x_0)}{b-a} \right) dx \ . \end{align*} Calculating the integral we have: \begin{align*} \int_a^b \exp\left( \frac{\mu (x-x_0)}{2D} \right) &\cos\left( \frac{n \pi (x-x_0)}{b-a} \right) dx = \\ \frac{D}{\lambda_n} \Bigg( \frac{\mu}{2D} & \left( \exp\left(\frac{\mu(b-x_0)}{2D} \right) \cos\left( \frac{n \pi (b-x_0)}{b-a} \right) - \exp\left(\frac{\mu(a-x_0)}{2D} \right) \cos\left( \frac{n \pi (a-x_0)}{b-a} \right) \right) \\ + \frac{n \pi}{b-a} & \left( \exp\left(\frac{\mu(b-x_0)}{2D} \right) \sin\left( \frac{n \pi (b-x_0)}{b-a} \right) - \exp\left(\frac{\mu(a-x_0)}{2D} \right) \sin\left( \frac{n \pi (a-x_0)}{b-a} \right) \right) \Bigg)\ . \end{align*} By inserting this in the expression for $S(t)$ we get: \begin{align*} S(t) &= \frac{2}{b-a} \exp\left(\frac{\mu(b-x_0)}{2D} \right) \sum_{n \in \mathbb{N}_0} \frac{D}{\lambda_n} \frac{\mu}{2D} \cos\left( \frac{n \pi (b-x_0)}{b-a} \right) \exp\big(-\lambda_n(t-t_0) \big) \\ &- \frac{2}{b-a} \exp\left(\frac{\mu(a-x_0)}{2D} \right) \sum_{n \in \mathbb{N}_0} \frac{D}{\lambda_n} \frac{\mu}{2D} \cos\left( \frac{n \pi (a-x_0)}{b-a} \right) \exp\big(-\lambda_n(t-t_0) \big) \\ &+ \frac{2}{b-a} \exp\left(\frac{\mu(b-x_0)}{2D} \right) \sum_{n \in \mathbb{N}_0} \frac{D}{\lambda_n} \frac{n \pi}{b-a} \sin\left( \frac{n \pi (b-x_0)}{b-a} \right) \exp\big(-\lambda_n(t-t_0) \big) \\ &- \frac{2}{b-a} \exp\left(\frac{\mu(a-x_0)}{2D} \right) \sum_{n \in \mathbb{N}_0} \frac{D}{\lambda_n} \frac{n \pi}{b-a} \sin\left( \frac{n \pi (a-x_0)}{b-a} \right) \exp\big(-\lambda_n(t-t_0) \big) \ . \end{align*} Now the first hitting time density $f_{\tau}(t)$ can be found by the expression: \begin{align*} f_{\tau}(t) = - \frac{dS}{dt} \ . \end{align*} $f_\tau$ is also known as the first passage time density. we see that: \begin{align*} f_{\tau}(t) &= \frac{2}{b-a} \exp\left(\frac{\mu(b-x_0)}{2D} \right) \sum_{n \in \mathbb{N}_0} \frac{\mu}{2} \cos\left( \frac{n \pi (b-x_0)}{b-a} \right) \exp\big(-\lambda_n(t-t_0) \big) \\ &- \frac{2}{b-a} \exp\left(\frac{\mu(a-x_0)}{2D} \right) \sum_{n \in \mathbb{N}_0} \frac{\mu}{2} \cos\left( \frac{n \pi (a-x_0)}{b-a} \right) \exp\big(-\lambda_n(t-t_0) \big) \\ &+ \frac{2}{b-a} \exp\left(\frac{\mu(b-x_0)}{2D} \right) \sum_{n \in \mathbb{N}_0} D \frac{n \pi}{b-a} \sin\left( \frac{n \pi (b-x_0)}{b-a} \right) \exp\big(-\lambda_n(t-t_0) \big) \\ &- \frac{2}{b-a} \exp\left(\frac{\mu(a-x_0)}{2D} \right) \sum_{n \in \mathbb{N}_0} D \frac{n \pi}{b-a} \sin\left( \frac{n \pi (a-x_0)}{b-a} \right) \exp\big(-\lambda_n(t-t_0) \big) \ . \end{align*} For convenience we define: \begin{align*} v(x,t) = \frac{2}{b-a} \exp\left(\frac{\mu(x-x_0)}{2D} \right) \sum_{n \in \mathbb{N}_0} \frac{n \pi}{b-a} \sin\left( \frac{n \pi (x-x_0)}{b-a} \right) \exp\big(-\lambda_n(t-t_0) \big) \ . \end{align*} Then we have: \begin{align*} f_{\tau}(t) &= \frac{\mu}{2} \Big( u(b,t) - u(a,t) \Big) + D \Big( v(b,t) - v(a,t) \Big) \ . \end{align*} By our boundary condition $u(b,t) = u(a,t) = 0$ thus: \begin{align*} f_{\tau}(t) &= D \Big( v(b,t) - v(a,t) \Big) \ . \end{align*}
The following plots are for $a=-1$, $b=1$, $x_0=0$ and $t_0 = 0$. In the plots we have $u:=\mu$.