It looks like I have been able to improve the previous partial affirmative answer to a complete one now:

Let $a:=\alpha$. 
Let $h:=f'' - 2a f' + 2a f$, so that $h\le0$. 
Without loss of generality (wlog), 
\begin{equation}
	\text{$f(x)\ge e^x - 1>0$ for all $x>0$.} \tag{1}
\end{equation} 

Let 
\begin{equation}
	r(x):=f(x)/e^{ax}. \tag{2}
\end{equation}
Then the equation $h=f'' - 2a f' + 2a f$ can be rewritten as 
\begin{equation*}
	r''+b^2r=g,\quad\text{where}\ g(x):=h(x)/e^{ax}\le0  
\end{equation*}
and $b:=\sqrt{(2-a)a}>0.$
In view of (1) and (2), $r>0$ (on $(0,\infty)$). So, $r''=-b^2r+g<0$, so that $r$ is a strictly positive concave function (on $(0,\infty)$). So, $r(x)\to R$ and $r'(x)\downarrow k$ as $x\to\infty$, for some $R\in(0,\infty]$ and $k\in[0,\infty)$. 

Take now any $\rho\in(0,R)$ and let 
\begin{equation*}
	d:=r-s,\quad\text{where }s(x):=\rho\sin bx. 
\end{equation*}
Then $d>0$ on $[x_*,\infty)$ for some real $x_*>0$. Also, $d''+b^2d=g$, so that $d''=-b^2d+g<0$ and hence $d$ is a positive concave function on $[x_*,\infty)$, so that $d'(x)\downarrow \ell$ as $x\to\infty$, for some $\ell\in[0,\infty)$. 
Thus, 
$b\rho\cos bx=s'(x)=r'(x)-d'(x)\to k-\ell$ as $x\to\infty$, which is absurd. Thus, the first inequality in (1) is false for some $x>0$, as desired.  

Actually, more than that is proved: we proved that $f(x_n)\le0$ for some sequence $x_n\to\infty$. Moreover, the initial conditions, $f(0) = 0$ and $f'(0) = 1$, have not been used or needed.