Conditional expectation of independent variables

In addition to the identical distribution independence is used to obtain that $(Z_1,Z_1+Z_2)$ and $(Z_2,Z_1+Z_2)$ are identically distributed which is needed to conclude $\mathbb E(Z_1 \mid Z_1+Z_2) = \mathbb E(Z_2 \mid Z_1+Z_2)$.

Edit. The distribution of $(Z_1,Z_1+Z_2)$ is the image (push forward) of $\mathbb P^{(Z_1,Z_2)}= \mathbb P^{Z_1} \otimes \mathbb P^{Z_2}$ (here independence is used) under the transformation $(u,v)\mapsto (u+v,v)$.

We need only show that, for any Borel set $A \in \mathbb{R}$, \begin{align*} \int_{Z_1+Z_2 \in A} Z_1 dP = \int_{Z_1+Z_2 \in A} Z_2 dP. \end{align*} We denote by $F$ the common cumulative distribution function of $Z_1$ and $Z_2$. Then, from the independence assumption, \begin{align*} \int_{Z_1+Z_2 \in A} Z_1 dP &= E\left(Z_1 \pmb{1}_{Z_1+Z_2 \in A} \right)\\ &=\iint_{\mathbb{R}^2} x \pmb{1}_{x+y \in A} dF(x) dF(y). \end{align*} Analogously, \begin{align*} \int_{Z_1+Z_2 \in A} Z_2 dP &= E\left(Z_2 \pmb{1}_{Z_1+Z_2 \in A} \right)\\ &=\iint_{\mathbb{R}^2} u \pmb{1}_{v+u \in A} dF(u) dF(v)\\ &=\iint_{\mathbb{R}^2} u \pmb{1}_{u+v \in A} dF(u) dF(v). \end{align*} That is, \begin{align*} \int_{Z_1+Z_2 \in A} Z_1 dP = \int_{Z_1+Z_2 \in A} Z_2 dP. \end{align*} In other words, \begin{align*} E\left(Z_1 \mid Z_1+Z_2 \right) = E\left(Z_2 \mid Z_1+Z_2 \right). \end{align*}