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Regarding a "rigorous but simple proof" of the relation the OP is interested in, such a proof is, almost completely, already written in the original post.

To see this, consider independent integrable random variables $X$ and $Y$ and assume that their characteristic functions, defined for every real number $t$, are such that $E({\mathrm e}^{\mathrm{i}tX})=\mathrm{e}^{a\psi(t)}$ and $E(\mathrm{e}^{\mathrm{i}tY})=\mathrm{e}^{b\psi(t)}$ for a given function $\psi$ and given real numbers $a$ and $b$. Let $S=X+Y$. Now, to prove that $(a+b)E(X\vert S)=aS$$ (a+b)E(X\vert S)=aS, $$ it suffices to show that$(a+b)E(X\mathrm{e}^{\mathrm{i}tS})=aE(S\mathrm{e}^{\mathrm{i}tS})$ , for every real number $t$. t$, $$ (a+b)E(X\mathrm{e}^{\mathrm{i}tS})=aE(S\mathrm{e}^{\mathrm{i}tS}). $$ Since both sides of the equality can be explicitly written in terms of $a$, $b$, the function $\psi$ and its derivative $\psi'$, the proof is, in a way and modulo some easy computations, already over.

For example, $$ E(S\mathrm{e}^{\mathrm{i}tS})=E(X\mathrm{e}^{\mathrm{i}tX})E({\mathrm e}^{\mathrm{i}tY})+E(Y\mathrm{e}^{\mathrm{i}tY})E({\mathrm e}^{\mathrm{i}tX})$ e}^{\mathrm{i}tX}), $$ because $X$ and $Y$ are independent. Here, both $E({\mathrm e}^{\mathrm{i}tX})$ and $E({\mathrm e}^{\mathrm{i}tY})$ are already known, and both $E(X{\mathrm e}^{\mathrm{i}tX})$ and $E(Y{\mathrm e}^{\mathrm{i}tY})$ are derivatives of the former with respect to $(\mathrm{i}t)$. Hence, $E(S\mathrm{e}^{\mathrm{i}tS})=-\mathrm{i}(a+b)\psi'(t)\mathrm{e}^{(a+b)\psi(t)}$.$ E(S\mathrm{e}^{\mathrm{i}tS})=-\mathrm{i}(a+b)\psi'(t)\mathrm{e}^{(a+b)\psi(t)}. $$ Likewise, $$ E(X\mathrm{e}^{\mathrm{i}tS})=E(X\mathrm{e}^{\mathrm{i}tX})E({\mathrm e}^{\mathrm{i}tY})=-\mathrm{i}a\psi'(t)\mathrm{e}^{(a+b)\psi(t)}$e}^{\mathrm{i}tY})=-\mathrm{i}a\psi'(t)\mathrm{e}^{(a+b)\psi(t)}. $$ Comparing these two formulas, we are done.

(If this helps, one can note that the signs of $a$ and $b$ must be the same, in the sense that $ab>0$ or that $X$ or $Y$ must be $0$ with full probability.)
show/hide this revision's text 2 added 82 characters in body

Regarding a "rigorous but simple proof" of the relation the OP is interested in, such a proof is, almost completely, already written in the original post.

To see this, consider independent integrable random variables $X$ and $Y$ and assume that their characteristic functions, defined for every real number $t$, are such that $E({\mathrm e}^{\mathrm{i}tX})=\mathrm{e}^{a\psi(t)}$ and $E(\mathrm{e}^{\mathrm{i}tY})=\mathrm{e}^{b\psi(t)}$ for a given function $\psi$. Hence the signs of $a$ and $b$ are the same \psi$ and , unless given real numbers $a=b=0$, one can (a$ and we will) assume that $a+b\ne0$. b$. Let $S=X+Y$ and $c=a/(a+b)$. It happens thatS=X+Y$. Now, to prove that $E(X\vert S)=cS$(a+b)E(X\vert S)=aS$, it suffices to show that $E(X\mathrm{e}^{\mathrm{i}tS})=cE(S\mathrm{e}^{\mathrm{i}tS})$ (a+b)E(X\mathrm{e}^{\mathrm{i}tS})=aE(S\mathrm{e}^{\mathrm{i}tS})$ for every real number $t$. Since both sides of the equality can be explicitly written in terms of $a$, $b$, the function $\psi$ and its derivative $\psi'$, the proof is, in a way and modulo some easy computations, already over.

For example, $E(S\mathrm{e}^{\mathrm{i}tS})=E(X\mathrm{e}^{\mathrm{i}tX})E({\mathrm e}^{\mathrm{i}tY})+E(Y\mathrm{e}^{\mathrm{i}tY})E({\mathrm e}^{\mathrm{i}tX})$ because $X$ and $Y$ are independent. Here, both $E({\mathrm e}^{\mathrm{i}tX})$ and $E({\mathrm e}^{\mathrm{i}tY})$ are already known, and both $E(X{\mathrm e}^{\mathrm{i}tX})$ and $E(Y{\mathrm e}^{\mathrm{i}tY})$ are derivatives of the former with respect to $\mathrm{i}t$. (\mathrm{i}t)$. Hence, $E(S\mathrm{e}^{\mathrm{i}tS})=-\mathrm{i}(a+b)\psi'(t)\mathrm{e}^{(a+b)\psi(t)}$.

Likewise, $E(X\mathrm{e}^{\mathrm{i}tS})=E(X\mathrm{e}^{\mathrm{i}tX})E({\mathrm e}^{\mathrm{i}tY})=-\mathrm{i}a\psi'(t)\mathrm{e}^{(a+b)\psi(t)}$. Comparing these two formulas, we are done.

(If this helps, one can note that the signs of $a$ and $b$ must be the same, in the sense that $ab>0$ or that $X$ or $Y$ must be $0$ with full probability.)
show/hide this revision's text 1

Regarding a "rigorous but simple proof" of the relation the OP is interested in, such a proof is, almost completely, already written in the original post.

To see this, consider independent integrable random variables $X$ and $Y$ and assume that their characteristic functions, defined for every real number $t$, are such that $E({\mathrm e}^{\mathrm{i}tX})=\mathrm{e}^{a\psi(t)}$ and $E(\mathrm{e}^{\mathrm{i}tY})=\mathrm{e}^{b\psi(t)}$ for a given function $\psi$. Hence the signs of $a$ and $b$ are the same and, unless $a=b=0$, one can (and we will) assume that $a+b\ne0$. Let $S=X+Y$ and $c=a/(a+b)$. It happens that, to prove that $E(X\vert S)=cS$, it suffices to show that $E(X\mathrm{e}^{\mathrm{i}tS})=cE(S\mathrm{e}^{\mathrm{i}tS})$ for every real number $t$. Since both sides of the equality can be explicitly written in terms of $a$, $b$, the function $\psi$ and its derivative $\psi'$, the proof is, modulo some easy computations, already over.

For example, $E(S\mathrm{e}^{\mathrm{i}tS})=E(X\mathrm{e}^{\mathrm{i}tX})E({\mathrm e}^{\mathrm{i}tY})+E(Y\mathrm{e}^{\mathrm{i}tY})E({\mathrm e}^{\mathrm{i}tX})$ because $X$ and $Y$ are independent. Here, both $E({\mathrm e}^{\mathrm{i}tX})$ and $E({\mathrm e}^{\mathrm{i}tY})$ are already known, and both $E(X{\mathrm e}^{\mathrm{i}tX})$ and $E(Y{\mathrm e}^{\mathrm{i}tY})$ are derivatives of the former with respect to $\mathrm{i}t$. Hence $E(S\mathrm{e}^{\mathrm{i}tS})=-\mathrm{i}(a+b)\psi'(t)\mathrm{e}^{(a+b)\psi(t)}$.

Likewise, $E(X\mathrm{e}^{\mathrm{i}tS})=E(X\mathrm{e}^{\mathrm{i}tX})E({\mathrm e}^{\mathrm{i}tY})=-\mathrm{i}a\psi'(t)\mathrm{e}^{(a+b)\psi(t)}$. Comparing these two formulas, we are done.