Let $X_1, X_2, X_3$ be independent RVs and $h: \mathbb{R} \to \mathbb{R}$ be measurable.
Claim: $$\mathbb{E}[h(X_1,X_2)\mid X_1, X_3] = \mathbb{E}[h(X_1,X_2)\mid X_1].$$
Why? Since $\sigma(X_1) \subset \sigma(X_1, X_3)$, I tried utilizing the tower property of CE. But due to a lack of measurability of either $\mathbb{E}[h(X_1,X_2)\mid X_1,X_3]$ w.r.t. $\sigma(X_1)$ or $h(X_1,X_2)$ w.r.t. $\sigma(X_1, X_3)$, I didn't really know how to proceed from $$ \mathbb{E} \big[ \mathbb{E}[h(X_1,X_2)\mid X_1,X_3]\mid X_1 \big] = \mathbb{E}[h(X_1,X_2)\mid X_1] = \mathbb{E} \big[ \mathbb{E}[h(X_1,X_2)\mid X_1]\mid X_1,X_3 \big].$$
$\def\B{\mathscr{B}}\def\R{\mathbb{R}}\def\peq{\mathrel{\phantom{=}}{}}\def\d{\mathrm{d}}\def\Π{{\mit Π}}$Without loss of generality, assmue that $X_1, X_2, X_3$ are real-valued random variables. For any $A, B \in \B(\R)$, the independence of $σ(X_1, X_2)$ and $σ(X_3)$ implies that\begin{align*} &\peq \int\limits_{\{X_1 \in A, X_3 \in B\}} E(h(X_1, X_2) \mid X_1) \,\d P\\ &= E\bigl(E(h(X_1, X_2) \mid X_1) I_A(X_1) I_B(X_3) \bigr)\\ &= E\bigl( E(h(X_1, X_2) \mid X_1) I_A(X_1) \bigr) E(I_B(X_3))\\ &= E\bigl(E(h(X_1, X_2) I_A(X_1) \mid X_1)\bigr) P(X_3 \in B)\\ &= E(h(X_1, X_2) I_A(X_1)) P(X_3 \in B), \end{align*}\begin{align*} &\peq \int\limits_{\{X_1 \in A, X_3 \in B\}} E(h(X_1, X_2) \mid X_1, X_3) \,\d P = \int\limits_{\{X_1 \in A, X_3 \in B\}} h(X_1, X_2) \,\d P\\ &= E(h(X_1, X_2) I_A(X_1) I_B(X_3)) = E(h(X_1, X_2) I_A(X_1)) E(I_B(X_3))\\ &= E(h(X_1, X_2) I_A(X_1)) P(X_3 \in B), \end{align*} thus$$ \int\limits_{\{X_1 \in A, X_3 \in B\}} E(h(X_1, X_2) \mid X_1, X_3) \,\d P = \int\limits_{\{X_1 \in A, X_3 \in B\}} E(h(X_1, X_2) \mid X_1) \,\d P. $$ Finally, note that $\Π := \bigl\{ \{X_1 \in A, X_3 \in B\} \mid A, B \in \B(\R) \bigr\}$ is a π-system and $σ(X_1, X_3) = σ(\Π)$, the identity follows from the π-λ theorem.