Does $E[X]\gg E[Y]$ for independent RV imply that $Pr[X+Y \geq x] \sim Pr[ X \geq x]$? We have two independent random variables $X$ and $Y$, where we know that
$E[X]\gg E[Y]$, thus $\frac{E[Y]}{E[X]}\rightarrow 0$. 
I am now interested in 
$Pr[X+Y \geq x]$ and would like to show that
$$Pr[X+Y \geq x] \sim Pr[X \geq x]$$
for $x=O(1)$.  
$X$ and $Y$ are both sums of independent Bernoulli random variables, thus $X=\sum_i X_i$ and $Y=\sum_j Y_j$ where $X_i, Y_j$ are independent. 
I tried to use the first moment method on $\frac{X}{Y}$ and also tried to extract the case where $Y=0$ but this all does not help. Would really appreciate if you could give me a hint, as small as it may be. 
To write
$$Pr[X+Y \geq x] = Pr[X \geq x] Pr[Y=0] + Pr[X=0] Pr[Y \geq x]+....$$
was my idea, but i don't see how this should help me. 

No.
Let $X=10$ be a constant and $Y$ take two values, $100$ and $-100$, with equal probabilities. Then $E(X)=10\gg 0=E(Y)$, but $P(X>15)=0$ while $P(X+Y)>15=1/2$.
I don't think you can get a "universal" inequality for all $x$. 
Take $x$ such that $P(X>x)$ is small but $P(Y>x)$ is large.
This does not contradict your condition on the expectations as is easily seen from the two examples you got.
What you would probably need is something like $P(X\gg Y)\approx 1$.
Note that the probability of density of $X+Y$ is the Convolution of their individual probability densities (which is a more complicated formula that the one you are trying to use).
