Wavefunction interpretations in QM From two-slit electron-interference experiment we can infer that there is a wave $\psi(x,t)$ that can be associated with electron. The amplitude at some point is the sum of amplitudes
reaching that point and the intensity is $|\text{Amplitude}|^2=|\psi(x,t)|^2$. Intensity can naturally be interpreted as probability density for finding electron at location $x$ at any given time.
Then we know from Fourier theory that any smooth wave $\psi(x,t)$ that vanishes at infinity can be given in a following form: $$\psi(x,t)=\int _{-\infty} ^\infty \hat{\psi}(k,t)e^{ikx}dk.$$
We also know from Parseval's identy that $$\int _{-\infty} ^\infty |\psi(x,t)|^2dx=\int _{-\infty} ^\infty |\hat{\psi}(k,t)|^2dk=1.$$
Now my question is that can we infer that $|\hat{\psi}(k,t)|^2$ is a density function for momentum without resorting to Schrödinger equation? Is there any experiment that we can use to conclude?
Secondly, we can construct wave $\psi(x,t)$ in a sense of energy: $$\psi(x,t)=\int _{-\infty} ^\infty \hat{\psi}(x,\omega)e^{-i\omega t}d\omega.$$
How easy it is to "see" that $|\hat{\psi}(k,t)|^2$ is a density function for energy?
Reason for my question is that if we have to those probability densities, then it would be easy to derive Schrödinger equation from one-dimensional fundamental lemma of calculus of variations.
 A: 
Now I just begin with double-slit experiment and then conclude that there is a probability wave for electron

This is wrong. Firstly, the wave function is defined on configuration space, it isn't a wave in space. And secondly, it isn't using probability the way a mathematics textbook on probability uses the word probability. So it isn't a wave in space, and it isn't a probability. A wave function is one (of many) possible mathematical descriptions of a physical system.

Now my question is that can we infer that $|\hat{\psi}(k,t)|^2$ is a density function for momentum without resorting to Schrödinger equation?

We can't infer it because it isn't true. It is not the case that $|\hat{\psi}(k,t)|^2$ is a probability density for momentum (or for wave number) and that $|{\psi}(x,t)|^2$ is a probability density for position.
In particular, there isn't a sample space on which two random variables (position and momentum) are defined and for which the values of both variables are determined by a region in the sample space, which is then unaffected by an observation/measurement. To do so would fundamentally deny the fact that the operators don't commute.
What we have sounds very similar but is different. We have a state. If you prepared many systems with that same state and measured position for some of them then the frequency of getting results within a region of width $\Delta x$ and containing some $x_0$ would approach $|{\psi}(x_0,t)|^2\Delta x$ as the number sampled gets large and $\Delta x$ gets small. This is different than a random variable on a sample space having a probability for different results. But for those states where you choose to measure the system you could make such a sample space.
Similarly, If you took those many systems with that same state and took the ones that you did not measure position on. Then if you "measure" momentum on those systems the frequency of getting results within a region of width $\Delta k$ and containing some $k_0$ would approach $|\hat{\psi}(k_0,t)|^2\Delta k$ as the number sampled gets large and $\Delta k$ gets small. This is different than a random variable on a sample space having a probability for different results. But for those states where you choose to measure the system you could make such a sample space.
So to be clear. If you pick up a text on probability they will start with a sample space, then have random variables and probabilities and then you do experiments. Here, you can't get a sample space until after you say what experiment you are going to do.  It might not seem very different, but when you do something even slightly different than what a mathematician assumed you were going to do, then their conclusions can't necessarily be applied.
So back to physics. You have a state, which is a mathematical description of the physical system. You have $|{\psi}(x,t)|^2$ which can be used to tell you the relative frequency of getting results of two different positions amongst the many identically prepared systems of which you choose to measure position. And you have $|\hat{\psi}(k,t)|^2$ which can be used to tell you the relative frequency of getting results of two different momentums amongst the many identically prepared systems of which you choose to measure momentum. so you have three mathematical descriptions of three different things. Two describe frequencies of (different and incompatible) experimental outcomes, one one describes the physical system.
If you tried to get a joint probability distribution for position and momentum, you'd have to use a Wigner type distribution which would be negative in some regions of the 6d position-momentum phase space but when you integrate out all the positions from the Wigner distribution you get $|\hat{\psi}(k,t)|^2$ and when you integrate out all the momentums from the Wigner distribution you get $|{\psi}(x,t)|^2.$ And the integrating is just the standard way of getting a marginal from a joint distribution, except of course this distribution is not nonnegative
So the reality is you merely get frequencies of different results for different experiments and doing the experiments changes things. And one distribution ($|{\psi}(x,t)|^2$ or $|\hat{\psi}(k,t)|^2$) doesn't tell you the other one 
($|\hat{\psi}(k,t)|^2$ or $|{\psi}(x,t)|^2$).
When you have the state, you have described the system completely, and can thus get the frequencies of any experiment whatsoever. Knowing the frequencies of one type of experimental setup is not enough information to tell you about different experimental setups.

Is there any experiment that we can use to conclude?

No. There are states with different momentum distributions that have the same position distribution, so knowing the latter cannot possibly tell us the former.

If $|\psi(x,t)|^2$ is not a probability density for position and $|\hat{\psi}(k,t)|^2$ for momentum, then I will burn few of my QM texbooks. This must be that "quantum weirdness"

You use it in a similar manner in some situations. For instance to get the frequencies of certain results in repeated experiments. But there is no nonnegative joint distribution that has marginals like that.
Your textbooks use the word probability because they don't want to cover how to relate a frequency to averages and all that over again. It just isn't a probability space until after you select the experiment. And they do mention non commutativity, and they probably do say that that it is only a probability distribution for a fixed measurement. You just have to read carefully. And there just is no joint distribution and knowing the distribution of one experiment doesn't tell you the distribution for a different experiment.
