Some simple observations, including a partial affirmative answer:

(i) Let $a:=\alpha$. 
Let $h:=f'' - 2a f' + 2a f$, so that $h\le0$. Since $f(0) = 0$ and $f'(0) = 1$, it follows that $f''(0)=2a+h(0)<1$ if $a<1/2$, so that for small enough $x>0$ one has $f(x)<x+x^2/2 < e^x - 1$. This gives the partial affirmative answer in the case $a\in(0,1/2)$. So, without loss of generality $a\in[1/2,2)$. 

(ii) If $x>0$ is the point of a local minimum of $f$, then $f'(x)=0$ and $f''(x)\ge0$, whence 
$2af(x)=h(x)-f''(x)+2af'(x)\le0$, and so, $f(x)\le0<e^x-1$. Thus, without loss of generality $f$ has no local minima in $(0,\infty)$. 

(iii) One has the following explicit form for $f$: 
\begin{equation}
	f(x)=e^{ax}\frac{\sin bx}b+\int_0^x e^{at}\frac{\sin bt}b\, h(x-t)\,dt
\end{equation}
for $x\ge0$, where $b:=\sqrt{(2-a)a}$.