Let us actually show more than requested, namely, that $f(x_n)\le0$ for some sequence $(x_n)$ converging to $\infty$ and all natural $n$. Moreover, the initial conditions, $f(0) = 0$ and $f'(0) = 1$, will not be used or needed.

Indeed, suppose that, to the contrary, the statement in the first sentence of this answer is false. Then
\begin{equation}
\text{there is some real $x_*>0$ such that $f(x)>0$ for all $x>x_*$.} \tag{1}
\end{equation}
Let $a:=\alpha$.
Let $h:=f'' - 2a f' + 2a f$, so that $h\le0$.
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 $(x_*,\infty)$. So, $r''=-b^2r+g<0$, so that $r$ is a strictly positive concave function on $(x_*,\infty)$.
It follows that $r'\ge0$ and hence $r$ is nondecreasing on $(x_*,\infty)$; indeed, if $r'(x_{**})<0$ for some real $x_{**}>x_*$, then the concavity of $r$ implies $r(x)\le r(x_{**})+r'(x_{**})(x-x_{**})\to-\infty$ as $x\to\infty$, which contradicts the positivity of $r$.
So, $r(x)\uparrow R$ 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 $\ell\in[0,\infty)$.
Therefore,
$b\rho\cos bx=s'(x)=r'(x)-d'(x)\to k-\ell$ as $x\to\infty$, which is absurd. Thus, (1) is false, and hence the statement in the first sentence of this answer is true.