Let's say that we have a model:
$y + x = \epsilon$, $\epsilon \sim N(0, 1)$.
After observing a value for $y$, we can write $x$ as:
$x = \epsilon - y$
Since $y$ is just a constant, and $\epsilon$ is standard normally distributed, it seems that we can claim:
$\epsilon - y \sim N(-y, 1)$
and thus:
$p(x \mid y) = N(x; -y, 1)$ where $N$ is the normal pdf.
But this makes no sense as there is no mention of a prior for $x$ anywhere. So what went wrong with this "equational" reasoning argument?
There are two ways to interpret your statement $\epsilon \sim \mathcal N(0, 1)$. If $\epsilon \sim \mathcal N(0, 1)$ is taken to mean that $\epsilon$ is marginally $\mathcal N(0, 1)$ then your logic falls apart at concluding that $[\epsilon - y \mid Y = y] \sim \mathcal N(-y, 1)$. This is because there is no reason whatsover to conclude that $[\epsilon \mid Y = y] \sim \mathcal N(0, 1)$; the assumption is made about the marginal distribution of $\epsilon$, but if $\epsilon$ and $Y$ are dependent we cannot make such a strong conclusion about the conditional distribution of $[\epsilon \mid Y]$.
Now, on the other hand, if one interprets $\epsilon \sim \mathcal N(0, 1)$ to mean that $[\epsilon \mid Y = y] \sim \mathcal N(0, 1)$ then your logic works. This is a standard assumption in parametric linear regression, after relabeling the variables slightly; one assumes $Y = X\beta + \epsilon$ and assumes that $[\epsilon \mid X] \sim \mathcal N(0, \sigma)$, from which is follows that $[Y \mid X] \sim \mathcal N(X\beta, \sigma)$.
"Since y is just a constant..."..it is not a constant. It is a sample from a distribution. The errors follow!!
I don't see any problem in your reasoning, except you want a prior on $x$. Note that $p(x,y,\epsilon) = p(x|y,\epsilon)p(y)p(\epsilon)$.
