I finally worked out the way to do it. Here it is for those of you interested. Thank you Davide for your attempt. Although it was wrong it inspired my proof.
Proof:
Suppose $\exists x \in V \setminus H$. To see that we must have $V=E$, choose $y \in E$ arbitrary. Also, write $V=U+a$.
We will have to distinguish two cases:
In this case $f(x-y)=0$. Let $y_0 \in E$ be s.t. $f(y_0) \neq 0$. Then
$x-y=\underbrace{x-y+ \alpha \frac{y_0}{f(y_0)}}_{\in H \subset V} - \underbrace{\alpha \frac{y_0}{f(y_0)}}_{\in H \subset V}$
So we can write
$x-y=u_0+a - (u_1+a) \implies y=\underbrace{x}_{\in V} + \underbrace{u_1-u_0}_{\in U} \in V $
In this case $f(x-y) \neq 0$. Then $\exists \beta \in \mathbb{R}$ (take $\beta=\frac{\alpha-f(y)}{f(x)-f(y)}$) s.t.
$f( \beta x+(1-\beta)y)=\alpha$
i.e. $\beta x+(1-\beta)y \in H \subset V$
$\implies \beta (a+u_0) +(1-\beta)y=a+u_1$
$\implies (1-\beta)y= (1-\beta) a + u_1-\beta u_0$
$\implies y= a + \underbrace{\frac{u_1-\beta u_0}{1-\beta}}_{\in U} \in V$
Hence, we conclude that $y\in V$ and since $y$ was arbitrary this implies that $V=E$.
$\square$