This is not an answer, but an unsuccessful attempt to reconcile the intuition of d'Alembert's formula with the explanation you've given, as well as an iteration of a cry for help since in my humble opinion there are same gaps in the intuition, which your question calls upon.
The general solution is
$$2u(x,t)=f(x-t)+f(x+t)+\int_{x-t}^{x+t}g(\tau)d\tau.$$
As you pointed out, if $g\equiv0$ then the initial wave $f$ splits in two and the solution goes in opposite directions (I imagine a string stretched and let go from rest). This case makes sense.
For the other situation, if $f\equiv0$, then for any $x$, $u(x,t)$ approaches the integral of $g$ as $t$ goes to infinity. This makes no intuitive sense, since straight from the formula we know then that if say, $g$ is any bump function with area 1, then for any $x$, $u(x,t)$ is eventually $1$, which is like saying "if you pluck a string with unit velocity, the entire string will eventually sit one unit in the direction that you've plucked it". This statement of course makes no sense physically...
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
$$u_{tt} - u_{xx} = 0$$
The general solution (without boundary condition) is : $$u(x,y)=f(x+t)+g(x+t)$$ The functions $f$ and $g$ are not necessarily the same.
The condition $u(x,0) = \phi(x)=f(x)+g(x)$ implies $\quad\begin{cases} f(x)=\frac12\phi(x)+h(x) \\ g(x)=\frac12\phi(x)-h(x) \end{cases}$
$$ u(x,t)=\frac12\phi(x+t)+h(x+t)+\frac12\phi(x-t)-h(x-t)$$
$h(x)$ is an arbitrary function to be determined by the condition $u_t(x,0)=0$
$$ u_t(x,t)=\frac12\phi'(x+t)+h'(x+t)-\frac12\phi'(x-t)+h'(x-t)$$ $$ u_t(x,0)=\frac12\phi'(x)+h'(x)-\frac12\phi'(x)+h'(x)$$ $$u_t(x,0)=0=2h'(x) \quad\implies\quad h(x)=C$$ Your solution is confirmed : $$ u(x,t)=\frac12\phi(x+t)+\frac12\phi(x-t)$$ The function $\phi$ is a given piecewise function : $$\phi(x) = \begin{cases} 1 & |x| \leq 1 \\ 0 & |x| > 1 \\ \end{cases}$$
$$ u(x,t)=\frac12\begin{cases} 1 & x+t \leq 1 \\ 1 & -x-t \leq 1 \\ 0 & x+t > 1 \\ 0 & -x-t > 1 \\ \end{cases}+ \frac12\begin{cases} 1 & x-t \leq 1 \\ 1 & -x+t \leq 1 \\ 0 & x-t > 1 \\ 0 & -x+t > 1 \\ \end{cases}$$
$$ u(x,t)=\frac12\begin{cases} 1 & x \leq 1-t \\ 1 & x \geq -1-t \\ 0 & x > 1-t \\ 0 & x < -1-t \\ \end{cases}+ \frac12\begin{cases} 1 & x \leq 1+t \\ 1 & x \geq -1+t \\ 0 & x > 1+t \\ 0 & x < -1+t \\ \end{cases}$$
We have to consider several regions limited by
$x=1-t \quad;\quad x=1+t \quad;\quad x=-1-t \quad;\quad x=-1+t\quad$
For $\quad \boxed{0<t<1}$ :
Case $\quad x>1+t \:: \quad u=\frac12(0)+\frac12(0)=0$.
Case $\quad 1-t<x\leq 1+t \: : \quad u=\frac12(0)+\frac12(1)=\frac12$.
Case $\quad -1+t<x\leq 1-t \: : \quad u=\frac12(1)+\frac12(1)=1$.
Case $\quad -1-t\leq x< -1+t \: : \quad u=\frac12(1)+\frac12(0)=\frac12$.
Case $\quad x\leq -1-t \: : \quad u=\frac12(0)+\frac12(0)=0$.
$$u(x,t)=\begin{cases} 0 && x>1+t \\ \frac12 && 1-t<x\leq 1+t \\ 1 && -1+t<x\leq 1-t\\ \frac12 && -1-t\leq x< -1+t\\ 0 && x\leq -1-t \end{cases}$$
The above formulas are not valid for $t>1$ which is outside the domain of study specified in the wording of the question.
IN ADDITION :
For $\quad \boxed{t>1}$ :
$$u(x,t)=\begin{cases} 0 && x>1+t \\ \frac12 && -1+t<x\leq 1+t \\ 0 && 1-t<x\leq -1+t\\ \frac12 && -1-t\leq x< 1-t\\ 0 && x\leq -1-t \end{cases}$$