Suppose $M(\lambda)$ is a distribution that satisfies for all $Z\sim M(\lambda)$ that $E[Z]=\lambda$. Does it follow that if $X \sim M(Y)$ then $E[X | Y] = Y$ a.s.? Intuitively this seems to be the case, but how does one prove it, if it is true?
Since this has been cause of a lot of confusion; some clarification:
Fix a probability space $(\Omega,\mathcal F, P)$ so that it can accommodate the following random variables. There is only this probability space involved.
$Y$ is a random variable with expectation that takes values in $\mathbb R$ and for every $\lambda\in\mathbb R$, $M(\lambda)$ is a distribution on $\mathbb R$.
Suppose $Q_\lambda$ is the measure with distribution $M(\lambda)$. I am assuming $Q_\lambda$ to be defined on the Borel sets of $\mathbb R$. I am defining $X\sim M(Y)$ to mean that $E[1_{\{X\in A\}}|Y] = P(X\in A | Y) = Q_Y(A)$ for all measurable $A\subseteq\mathbb R$ $P$-a.s.
The conditional expectation $E[X | Y]$ is then defined as usual, as the a.s. uniquely determined random variable $H$ satisfying $E[ZH] = E[ZX]$ for all $\sigma(Y)$-measurable bounded $Z$.
Best Answer
You say that for each $\lambda\in\mathbb{R}$, $M(\lambda)$ is a distribution on $\mathbb{R}$, by which I assume you mean a Borel probability measure. Let $\mathcal{R}$ denote the Borel $\sigma$-algebra on $\mathbb{R}$. Let us further suppose that for every $A\in\mathcal{R}$, the function $M(\cdot,A)$ is Borel measurable. Then $M$ is what's called a probability kernel from $\mathbb{R}$ to $\mathbb{R}$. It follows that $M(Y)$ is a random measure, meaning $M(Y(\omega))$ is a measure for each $\omega$ and $M(Y(\cdot), A)$ is a random variable for each $A$. It can be show that this implies $M(Y)$ is a measurable function from $\Omega$ to $M_1(\mathbb{R})$, the space of Borel probability measures on $\mathbb{R}$, where we equip $M_1$ with the $\sigma$-algebra generated by the projection maps, $\sigma(\{\pi_A: A \in \mathcal{R}\})$. (Here, $\pi_A$ is the map $\nu\mapsto \nu(A)$.)
A regular conditional distribution for $X$ given $\mathcal{G}$ is a random measure $\mu$ such that $P(X\in A\mid \mathcal{G})=\mu(\cdot,A)$ a.s. for each $A$. Your hypothesis assert that $M(Y)$ is a regular conditional distribution for $X$ given $Y$ (strictly speaking, given $\sigma(Y)$).
A special case of Theorem 5.4 in Foundations of Modern Probability by Olav Kallenberg states that if $\mu$ is a regular conditional distribution for $X$ given $Y$ and $f:\mathbb{R}^2\to\mathbb{R}$ is measurable with $E|f(X,Y)|<\infty$, then $$ E[f(X,Y) \mid Y](\omega) = \int_\mathbb{R} f(x,Y)\,\mu(\omega,dx) $$ for $P$-almost every $\omega\in\Omega$. This can be proven as follows. If $f=1_{A\times B}$, it follows easily from the hypotheses. Use the $\pi$-$\lambda$ theorem to prove it for all indicators $f=1_C$. Use linearity of conditional expectation to prove it for all simple $f$. Use the monotone convergence theorem for conditional expectations to prove it for all nonnegative $f$. Then apply this to the positive and negative parts of $f$.
In your case, since $\int x\,M(y,dx)=y$ for all $y$, the above yields $$ E[X \mid Y] = \int_\mathbb{R} x \, M(Y,dx) = Y \text{ a.s.,} $$ which is what you were hoping to conclude.