Let $f:\Omega \rightarrow \mathbb{C}$ be a holomorphic function where $\Omega:=\{z\in\mathbb{C}; -1<Re(z)<|Im(z)|\}$ (i.e. $\Omega$ is open, connected, non-convex). If
$|f'(z)-1|<1/4$
then $f$ is injective on $\Omega$.
PS.: 1)It is obvious that it is gonna be locally injective since $f'(z)\neq 0$, $\forall z\in\Omega$ (using b)... but how can I guarantee that it is gonna be for whole $\Omega$?
2) If $\Omega$ is convex it is trivial:
In fact, suppose that $a\neq b$ s.t $f(a)=f(b)$. So, consider the straight line $\gamma$, connecting $a$ to $b$, i.e. $\gamma:[0,1]\rightarrow \Omega$ is defined as $\gamma (t)=a+(b-a)t$. Since $\Omega$ is convex, this line lies in $\Omega$ and so $f$ is holomorphic here. Thus,
$0=f(b)-f(a)=\displaystyle\int_{\gamma}f'(z)dz=\int_{0}^{1}f'(\gamma(t))\gamma'(t)dt=\int_{0}^{1}f'(a+(b-a)t)(b-a)dt$
As $(b-a)\in\mathbb{C}^{*}$ is just a constant (non-zero since $a\neq b$, by assumption). We have:
$\displaystyle0=\int_{0}^{1}f'(a+(b-a)t)dt$ $\Rightarrow \displaystyle -1=\int_{0}^{1}(f'(a+(b-a)t)-1)dt $ (adding (-1) both sides). Now, Taking the module, we have
$1=\displaystyle|\int_{0}^{1}(f'(a+(b-a)t)-1)dt|\leq \int_{0}^{1}|(f'(a+(b-a)t)-1)|dt<\int_{0}^{1}\frac{1}{4}dt=\frac{1}{4}$ (we have to use the assumption here for the last inequality). So, $1<\frac{1}{4}$, Contradiction. Therefore, $f(a)\neq f(b)$, i.e. $f$ is injective. $\square$
$\Omega$:
For your revised problem, note that there are various "large" convex subsets of $\Omega.$ In each of those $f$ is injective as we've seen. Play around a bit with these and you'll see that we can reduce the problem to this: If $a\in \Omega\cap \{y>|x|\},$ $b\in \Omega\cap \{y<-|x|\},$ then $f(a)\ne f(b).$
To prove this, let $a=a_1+ia_2, b=b_1+ib_2.$ Choose $0 < \epsilon < \min(1,a_2,|b_2|).$
Then
$$\tag 1 f(a)-f(-\epsilon) = \int_{[-\epsilon,a]}f'(z)\,dz = (a+\epsilon)\int_0^1 f'(-\epsilon + t(a+\epsilon))\,dt.$$
Now by the assumption on $f',$ the last integral has the form $A_1+iA_2,$ where $A_1>3/4$ and $|A_2|<1/4.$ We can write $(1)=(a_1+\epsilon +ia_2)(A_1+iA_2).$ The imaginary part of this is
$$ (a_1+\epsilon)A_2 + a_2A_1 >-2|a_1||A_2|+ a_2A_1 > -2a_2(1/4) + a_2(3/4) = a_2/4>0.$$
You can verify in a similar way that the imaginary part of $f(b)-f(-\epsilon)$ is less than $b_2/4<0.$ Thus
$$f(a)-f(b)=f(a) - f(-\epsilon) - (f(b)-f(-\epsilon))$$
has positive imaginary part. This shows $f(a)\ne f(b)$ as desired.
I don't see that we need a) in this problem.
Prvious answer: You've gone from a difficult question to an easy one with your edit. Here is a well known result: Suppose $\Omega$ is open and convex, and that $f$ is holomorphic in $\Omega,$ with $\text { Re } f' > 0$ in $\Omega.$ Then $f$ is injective. Proof: Just do what @ArnaudMortier did. Here his proof idea will work because you can take $\gamma$ to be a straight line segment. I'll leave the verification to you.
The assumptions on $f$ in your question are overkill. Your $\Omega$ is a vertical strip, hence is convex. And b) gives $\text { Re } f' > 0$ (and a lot more). We don't need a) at all.