Why are we not concerned about the distribution of the $x_t$ in an AR(1) model? I am trying to investigate the reasons why we don't bother about the distribution of the $x_t$ in an autoregressive model. 
Why do we concern ourselves about the distribution of $e_t$?
And why are the assumptions meant for $e_t$ alone knowing well that $e_t$ is as a result of $x_t-x_t^{estimated}$?
 A: Starting from $x_t-\hat x_t$ is not how we build the statistical theory around this model.
(This is also characteristic to a much broader class of statistical models than just the AR(1).)
The AR(1) model for a time series process $x_t$ specifies that 
$$
x_t=\varphi x_{t-1} + \varepsilon_t
$$
with $\varepsilon_t \sim i.i.d.(0,\sigma^2)$ and $\varepsilon_t$ being uncorrelated with $x_{t-1}$. Note that the distribution of $x_t$ is not specified in this definition. 
Given the definition, we can derive properties of $x_t$ as a process; properties of estimators of $\varphi$ and $\sigma^2$; and properties of some test statistics based on the latter estimators, e.g. a test that $H_0\colon\ \varphi=0$.


*

*To derive the maximum likelihood estimator, we need to know or assume the distribution of $\varepsilon_t$.  

*To derive the conditional least squares estimator we do not need a distributional assumption on $\varepsilon_t$.  

*Whatever the estimator (maximum likelihood, conditional least squares, ...), we need to know or assume the distribution of $\varepsilon_t$ to derive finite-sample properties of the estimator of $\varphi$ and test statistics based on the estimators.


In any of these cases, the distribution of $x_t$ is irrelevant for the statistical theory around the model. In the derivations of the maximum likelihood estimator or the test statistics, the distribution of $x_t$ either plays no role or if it does, it cancels out in the final result. So it is simply irrelevant.
P.S. Note that there is a difference between the error (or shock, or innovation) $\varepsilon_t$ and the model residual $\hat\varepsilon_t:=x_t-\hat x_t$. We make assumptions on the former, not the latter.
A: Next to the +1 answer by @RichardHardy, note that recursively solving for $x_{t-j}$ yields the so-called $MA(\infty)$-representation
\begin{eqnarray}
x_t&=&\epsilon_t+\varphi(\varphi x_{t-2}+\epsilon_{t-1}) \notag\\
&=&\epsilon_t+\varphi\epsilon_{t-1} +\varphi^2 x_{t-2}\notag\\
&=&\epsilon_t+\varphi\epsilon_{t-1}+\varphi^2\epsilon_{t-2} +\ldots+\varphi^j\epsilon_{t-j}+\varphi^{j+1} x_{t-j-1}
\end{eqnarray}
If $|\varphi|<1$ $\varphi^{j+1} x_{t-j-1}$ will vanish (in a suitable sense) as $j\to\infty$, such that
$$
x_t=\sum_{j=0}^\infty\varphi^j\epsilon_{t-j}
$$
Hence, $x_t$ is a function of present and past $\epsilon_t$, such that there is no need to specify any distributional properties of $x_t$ over and above those for (if any) $\epsilon_t$.
That said, even in "standard" regression models in which such recursive substitution of the regressors is not possible, we are typically interested in, for example, the conditional mean $E(y|X)$ so that we do not care about the marginal distribution of the regressors. 
