I'm going to assume you intended $\theta$ and $\varepsilon$ to be independent. Omission of that information is a frequent mistake.
Your proposal that the conditional expected value of $\theta$ given $\tilde x = x$ is $x$ would be right if $\tilde x$ and $\varepsilon$ were independent. But if $\theta$ and and $\varepsilon$ are independent, then the correlation between $\tilde x = \theta+\varepsilon$ and $\theta$ is positive, not zero.
Consider the case where the variance of $\theta$ is small and the variance of $\varepsilon$ is large: in that case, $\tilde x$ being a long way from $y$ would more likely result from variation of $\tilde x$ than from $\theta$ being far from $y$, so the conditional expected value of $\theta$ would certainly not be $\tilde x$.
One way to look at conditional distributions in this situation is via Bayes's theorem, which says the posterior distribution is proportional to the product of the prior distribution and the likelihood function. The prior distribution of $\theta$ is proportional to
$$
\sqrt a \exp\left( \frac{-1}2 a^2 (t-y)^2 \right) \,dt.
$$
The likelihood function given the observation that $\tilde x = x$ is
$$
L(t) = \sqrt b \exp\left( \frac{-1}2 b^2 (x - t)^2 \right).
$$
Notice that in the first line above you see $dt$, indicating that this is a measure on the $t$-axis, and no $dt$ appears in the second line; this is not a measure but a pointwise defined function. Multiplying a measure by a pointwise defined function yields a measure, so you expect to see $dt$ again.
To multiply these, we add the exponents:
\begin{align}
& a(t-y)^2 + b(x-t)^2 \\[10pt]
= {} & (a+b) t^2 - 2(ya+xb)t + \text{constant} \\
& \quad \text{(and in this case “constant" means not depending on $t$)} \\[10pt]
= {} & (a+b) \left( t^2 - 2 \frac{ya+xb}{a+b} t \right) + \text{a different constant} \\[10pt]
= {} & (a+b) \left( t - \frac{ya+xb}{a+b} \right)^2 + \text{yet another constant}.
\end{align}
We get the posterior distribution
$$
\text{constant} \times \exp\left( (a+b) \left( t - \frac{ya+xb}{a+b} \right)^2 \right) \, dt.
$$
Hence the posterior distribution is a normal distribution with variance $1/(a+b)$ and expected value is $(ya+xb)/(a+b).$
As a way to check this, find
$$
\operatorname{cov}\left( \tilde x, \theta - \frac{ay+b\tilde x}{a+b} \right),
$$
You should get $0$, so they're uncorrelated, and in a multivariate normal setting that means they're independent. If two random variables are independent, then the conditional expected value of the first one given the second, is the first one, so we have
$$
\operatorname{E}\left( \theta - \frac{ay+b\tilde x}{a+b} \mid \tilde x \right) = 0,
$$
and that implies
$$
\operatorname{E}\left( \theta \mid \tilde x \right) - \operatorname{E}\left( \frac{ay+b\tilde x}{a+b} \mid \tilde x \right) = 0.
$$
Since the second conditional expected value is determined by $\tilde x$, we have
$$
\operatorname{E}( \theta \mid \tilde x) - \frac{ay+b\tilde x}{a+b} = 0.
$$
