Bayes' Theorem versus my intuition for Y = X + Z Suppose $X$ and $Z$ are standard normal random variables and define $$Y = X + Z.$$
Then $Y$ is also normal, has zero mean and its variance is two (Wikipedia). It should also be fair to assert that $Y|(X=x)$ has a normal distribution with unit variance and mean $x$.
I apply Bayes' Theorem and find
$$
P_{X|Y}(x;y) = \frac{P_{Y|X}(y;x)P_X(x)}{P_Y(y)} = \frac{e^{-\frac{1}{4} (y-2 x)^2}}{\sqrt{\pi }}.
$$
My intuition is that $X|(Y=y)$ would be normally distributed with mean $y$ and unit variance. After all, $X=Y - Z$. Where've I gone wrong?
 A: 
Suppose $X$ and $Z$ are standard normal random variables and define $$Y = X + Z.$$
  Then $Y$ is also normal, has zero mean and its variance is two (Wikipedia).

You're failing to state some necessary assumptions.
Given only that $X$ and $Z$ are marginally standard normal, it does not follow that $X+Z$ is normal, nor does it follow that $V(X+Z)=V(X)+V(Z)$; however, the first conclusion does follow if $X$ and $Z$ are also assumed to be  independent, and the second conclusion does follow if $X$ and $Z$ are also assumed to be uncorrelated, i.e. $\text{Cov}(X,Z)=0$. (Of course they are uncorrelated if they are independent.)

It should also be fair to assert that $Y|(X=x)$ has a normal distribution with unit variance and mean $x$.

If we assume that $X$ and $Z\sim N(0,1)$ are independent, then your  assertion is easy to prove:
$$\begin{align}&P(Y\le y\mid X=x)\\
&=P(X+Z\le y\mid X=x)\\
&=P(Z\le y-x\mid X=x)\\
&=P(Z\le y-x)\quad\text{because }X,Z\text{ are independent}\\
&=\Phi(y-x)
\end{align}$$
which is the CDF of a $N(x,1)$ variate.

I apply Bayes' Theorem and find
  $$
P_{X|Y}(x;y) = \frac{P_{Y|X}(y;x)P_X(x)}{P_Y(y)} = \frac{e^{-\frac{1}{4} (y-2 x)^2}}{\sqrt{\pi }}.
$$

That's correct; i.e., $X|(Y=y) \sim \text{Normal}(\text{mean}=\color{blue}{\frac{y}{2}},\text{variance}=\color{red}{\frac{1}{2}})$:
$$P_{X|Y}(x;y) = \frac{ e^{-\frac{1}{2\color{red}{\frac{1}{2}}}(x-\color{blue}{\frac{y}{2}})^2} }{{\sqrt{2\pi\color{red}{\frac{1}{2}}}}}.
$$

My intuition is that $X|(Y=y)$ would be normally distributed with mean $y$ and unit variance. After all, $X=Y - Z$. Where've I gone wrong?

Your intuition is wrong, perhaps because you seem to be reasoning like this: $X=Y - Z$, therefore $X|(Y=y)=y-Z$, so $E(X|Y=y)=y-E(Z)=y-0=y$, etc. 
That's incorrect, because it uses a conditional expectation on the LHS but an uncondional expectation on the RHS. If done consistently with conditional expectations, we get $E(X\mid Y=y)=E(Y-Z\mid Y=y)=E(Y\mid Y=y)-E(Z\mid Y=y)=y-E(Z\mid Y=y)$. Now, because $X$ and $Z$ are identically distributed, they have the same conditional distribution given their sum $Y=X+Z=y$, so $E(Z\mid Y=y)=E(X\mid Y=y)$; hence  $E(X\mid Y=y)=y-E(X\mid Y=y)\implies E(X\mid Y=y)=\frac{y}{2}$, consistent with the Bayes Theorem result.
Similarly for the conditional variance:
$$\begin{align}&V(X\mid Y=y)\\
&= V(Y-Z\mid Y=y)\\
&= V(Y\mid Y=y) + V(-Z\mid Y=y) + 2\,\text{Cov}(Y,-Z\mid Y=y)\\
&= V(y\mid Y=y) + V(Z\mid Y=y) - 2\,\left(E[y\cdot Z\mid Y=y]-E[y\mid Y=y]\cdot E[Z\mid Y=y]\right)\\
&= 0 + V(Z\mid Y=y) - 2\,\left(y\cdot E[Z\mid Y=y]-y\cdot E[Z\mid Y=y]\right)\\
&= V(Z\mid Y=y) - 0\\
&= V(Z|Y=y)
\end{align}$$ 
which is simply a correct identity, again due to the fact that because $X$ and $Z$ are identically distributed, they have the same conditional distribution given their sum $Y=X+Z=y$.
A: After some thought, I can answer my own question. The key is using both definitions of $X$, which for clarity are
$$
\begin{align}
X &= Y - Z \\ 
X &= W,
\end{align}
$$
where $W$ is another standard normal random variable (this is just the original definition of $X$). Even given $Y=y$, the two equations are not inconsistent and simply give
$$
X|(Y=y) = \frac{1}{2} \left( y - Z + W \right).
$$
The result is a normal distribution with mean $y/2$ and a variance equal to $1/2$, which is also the result from Bayes' Theorem in my question. I wonder how you would approach a more complicated equation, say
$$
\begin{align}
Y &= f(X,Z) \\
X &= W,
\end{align}
$$
where you cannot solve for $X|(Y = y)$?
