I have two independent random variables $X,Y$ with the same following distributions : $$P(X=0)=P(Y=0)=1-p\,,\, P(X=1)=P(Y=1)=p$$
I want to calculate $E(X\mid 1_{\{X+Y=0\}})$.
So Let's define random variable $Z=1_{\{X+Y=0\}}$.
Then $P(Z=1)=(1-p)^2$ and $P(Z=0)=1-(1-p)^2$
$E(X\mid Z)=E(X\mid Z=0)P(Z=0)+E(X\mid Z=1)P(Z=1)$.
So we see that if $Z=0 \Rightarrow X=0 \Rightarrow E(X\mid Z=0).$ No now we only have to count $E(X\mid Z=1)$.
$E(X\mid Z=1)=E(X\mid \{X=0\} \cap\{Y=1\}) \cdot P(\{X=0\} \cap\{Y=1\})+E(X\mid \{X=1\} \cap\{Y=1\}) \cdot P(\{X=1\} \cap\{Y=1\})+E(X\mid \{X=1\} \cap\{Y=0\}) \cdot P(\{X=1\} \cap\{Y=0\})=p$.
Are Am i correctly with my transformations ?
No. $\mathbb E(X\mid Z)$ is a function of $Z$. The values of this function can be found assigning values to $Z$. Say, if we define a function $h$ on $\{0,\,1\}$ as $$h(x)=\mathbb E(X\mid Z=x) $$ then $\mathbb E(X\mid Z)=h(Z)$.
So we need to calculate $h(0)$ and $h(1)$.
\begin{align} h(0) = \mathbb E(X\mid Z=0) &= 0\cdot \mathbb P(X=0\mid Z=0)+1\cdot \mathbb P(X=1\mid Z=0) \\&= \mathbb P(X=1\mid X+Y\neq 0) \\&=\frac{\mathbb P(X=1)}{\mathbb P(X+Y\neq 0)} \\&=\frac{p}{1-(1-p)^2}=\frac{1}{2-p} \end{align}
and
\begin{align} h(1) = \mathbb E(X\mid Z=1) &= 0\cdot \mathbb P(X=0\mid Z=1)+1\cdot \mathbb P(X=1\mid Z=1) \\&= \mathbb P(X=1\mid X+Y =0)=0 \end{align}
So the answer is
\begin{align} \mathbb E(X\mid Z)=h(Z)&=\begin{cases}0 &, & Z=1 \cr \frac{1}{2-p} &, & Z=0 \end{cases} \\&= \frac{1}{2-p}1_{\{Z=0\}} \end{align}