First of all, you should distinguish the spacetime regions $x\le ct$ (boundary layer) and $x>ct$. When $x> ct$, the influence of the boundary is not felt, and the solution is zero due to the initial condition being zero. From now on assume $x\le ct$.
The formula
$$u(x,t) = \int_{0}^{x-ct} h(-s/c)ds + G(ct-x) + G(x+ct)$$
is correct, although the integral is awkwardly written: the upper limit is negative. I would introduce $H(t)=\int_0^t h(s)\,ds$ and write
$$
\int_{0}^{x-ct} h(-s/c)ds = -c \int_0^{t-x/c} h(\xi)\,d\xi = -c H(t-x/c)
$$
So far,
$$u(x,t) = -c H(t-x/c) + G(ct-x) + G(x+ct)\tag{1}$$
But we know that $G(ct-x), G(x+ct) =0$ from IC
Yes. Note that plugging $t=0$ here is problematic because we work in the regime $x\le ct$. It's better to plug $x=ct$, where (by continuity) $u=0$. So,
$$ 0 = G(0) + G(2x)$$
hence $G$ is identically zero.
The final answer:
$$
u(x,t) = \begin{cases} -cH(t-x/c),\quad &0\le x\le ct; \\ 0,\quad & x>ct\end{cases}
$$
Verification: in the boundary layer $x< ct$ we have $u_x(x,t) = h(t-x/c)$, which at $x=0$ gives $h(t)$ as required. The initial conditions are also satisfied, and the function is continuous.
It seems I was confused or made mistakes the first time trying to solve this, so I will post my solution now that it makes sense.
Assume $u(x,t)=F(x+2t)+G(x-2t)$. Then, the initial conditions give us:
$$
u(x,0)=F(x)+G(x)=e^{-x}\\
u_t(x,0)=2F'(x)-2G'(x)=2e^{-x}
$$
which hold for all $x>0$.
Hence, in the last equation, we may divide both sides by 2 and integrate with respect to x, then solve the resulting system:
$$
\begin{cases}
F(x)+G(x)=e^{-x}\\
F(x)-G(x)=-e^{-x}+C
\end{cases}
$$
where $C$ is some constant.
We thus obtain $F(x)=C/2$ and $G(x)=e^{-x}-C/2$. Since the only condition is that $x$ is positive, we may replace it by any positive quantity (I think this is where I was previously confused). Thus, we obtain $u(x,t)=F(x+2t)+G(x-2t)=C/2+e^{-(x-2t)}-C/2=e^{2t-x}$ which holds for all $x>2t$. It is clear, as I noted in my original post, that this corresponds to the same solution that we get from d'Alembert's formula.
Now we handle the case when the argument to $G$ is negative. For this, we need to use the boundary condition.
We have $u_x(0,t)=F'(2t)+G'(-2t)=-\cos t$ for all $t>0$. Make the substitution $z=-2t$ to obtain $G'(z)=-\cos(z/2)-F'(-z)=-\cos(z/2)$ by noting that $F'(-z)=0$ from our previous work. Integrating, we obtain $G(z)=-2\sin(z/2)+\tilde{C}$.
Applying the continuity condition, we must have $G(0)=\tilde{C}=1-C/2$. Thus, we have $u(x,t)=F(x+2t)+G(x-2t)=1-2\sin(x/2-t)$ for $0<x<2t$.
Hence, the complete solution is:
$$u(x,t)=\begin{cases}1-2\sin(x/2-t),&0<x\leq 2t\\e^{2t-x},&x>2t\end{cases}.$$
We can now verify that the initial conditions, boundary condition, and continuity are satisfied.
Best Answer
The condition given is not a Robin type of condition. Robin conditions involve the value of the function and the first derivative at the same endpoint. You have a Dirichlet and a Neumann condition, as well as an initial condition.
This problem can be solved by separation of variables. $$ \frac{T''(t)}{9T(t)} = \lambda, \;\; \lambda = \frac{X''(x)}{X(x)} $$ with spatial conditions $$ X'(0)=0,\;\; X(1)=0. $$ The eigenfunctions of this problem have the form $$ X_n(x) = \cos(\lambda_n x),\;\;\;\lambda_n = \frac{\pi}{2}+n\pi,\;\; n=0,1,2,\cdots. $$ The corresponding solutions $T_n$ are $$ T_n(t) = A_n\cos(3\lambda_n t)+B_n\sin(3\lambda_n t). $$ So the general solution is $$ u(t,x)=\sum_{n=1}^{\infty}(A_n\cos(3\lambda_n t)+B_n\sin(3\lambda_n t))\cos(\lambda_n x). $$ There are not enough conditions to uniquely solve the problem because the one $t=0$ condition does not involve the constants $B_n$. However, one solution is obtained by setting all $B_n=0$, which is equivalent to imposing the additional condition $u_t(0,x)=0$. The remaining condition is $$ 2\cos(3\pi x/2) = \sum_{n=1}^{\infty} A_n\cos(\lambda_n x) = \sum_{n=1}^{\infty}A_n\cos((n+1/2)\pi x). $$ The eigenfunctions $\{ \cos((n+1/2)\pi x) \}_{n=1}^{\infty}$ are mutually orthogonal on $[0,1]$, which allows you to isolate the $A_n$ using the above condition.