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 (wlog) $a\in[1/2,2)$. 

(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, wlog $f$ has no local minima in $(0,\infty)$. Since wlog 
\begin{equation}
	\text{$f(x)\ge e^x - 1$ for all $x\ge0$,} \tag{1}
\end{equation}
it further follows that wlog $f$ is increasing on $[0,\infty)$. 

(iia) One can considerably strengthen/generalize point (ii): assuming, wlog, that $f(x)>0$ for all $x>0$ (which follows from (1)), let us show that, for any real $c$,  $r(x):=f(x)/e^{cx}$ has no local minima in $(0,\infty)$; in particular, it will then follow from (1) that $f(x)/e^{cx}$ is increasing for any $c<1$. 
Indeed, suppose that $r$ has a local minimum at some real $x>0$. Then $r'(x)=0$ and $r''(x)\ge0$, which can be rewritten as $f'(x)=cf(x)$ and $0\le f''(x)-2cf'(x)+c^2f(x)=f''(x)-c^2f(x)$, so that $f''(x)\ge c^2f(x)$ and hence $0\ge h(x)=f''(x) - 2a f'(x) + 2a f(x)\ge[c^2+2a(1-c)]f(x)=[(c-a)^2+b^2]f(x)$, with $b:=\sqrt{(2-a)a}>0$. It follows that $f(x)\le0$, a contradiction. 

(iib) Quite similarly, one can show that (1) implies that a local minimum of $s(x):=\frac{f(x)}{e^x-1}$ may occur at $x>0$ only if $x\le\ln(2a)$. 

(iic) The same conclusion holds for $d(x):=f(x)-(e^x-1)$ instead of $s(x)$, and it is even easier to prove. Indeed, suppose that $d$ has a local minimum at some real $x>0$. Then $d'(x)=0$ and $d''(x)\ge0$, which can be rewritten as $f'(x)=e^x$ and $f''(x)\ge e^x$, which yields $0\ge h(x)=f''(x) - 2a f'(x) + 2a f(x)\ge e^x-2ae^x+2af(x)$, whence, in view of (1), $e^x - 1\le f(x)\le \frac{2a-1}{2a}\,e^x$, $e^x\le2a$, and $x\le\ln(2a)$. 

(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 $x\ge0$, with $b=\sqrt{(2-a)a}$ as before.