Instead of tedious derivations, simply invoke the invariance property of MLEs. Then solve for $x$ using basic algebra. However, note that this approach will lead to the same estimator you derived, i.e., $\hat{x}=-\log(2m/n -1)$.
So what to do? First, ignore estimation for the moment. Look at what the true value of $x$ would be if you knew the true value of $\theta$.
Suppose $\theta = 0$. Then, $e^{-x} = -1$, which implies that $x = - \log (-1)$. As you noted, this is not a real value.
So what about $\theta = 1/2$? We have that $e^{-x} = 0$, which is a problem for you yet again.
The point here is that with a couple of special cases based on known values, we can see that $x$ is undefined in a large number of cases. Your estimation is not wrong per se. The value of $x$ is simply undefined (not real) for certain values of $\theta$.
1) Meta-issue:
I believe the OP should include in the title of the question a signal that something special is going on here - for example instead of "Are linear combinations of..." the title should read "Are $\mathbb{F}_2-$linear combinations of...", so that the reader gets the message that concepts may have special meanings in this specific question.
2) Bernoulli or not Bernoulli?
By $\mathbb{F}_2-$arithmetic, the sum of i.i.d Bernoulli rv's is not a Binomial, since all probabilities are condensed on the values $0$ and $1$. We first derive the binomial distribution and then add the probabilities of all even values of the support to the probability of $0$, and the probabilities of all odd values of the support to the probability of $1$ -and thus we obtain again a Bernoulli r.v. This is just to validate the probability mass function included in the OP's question. It is by construction a Bernoulli random variable, irrespective of how $P(v'_i = 1)$ is derived. Moreover if the number of $1$'s is the same in each row of the matrix $A$, then each $v'_i$ has an identical Bernoulli marginal distribution.
3) $\mathbb{F}_2-$ linear independence.
$\mathbb{F}_2-$ linear dependence does not look much like the "usual" concept of linear independence. To be able to obtain a square matrix of full row/column rank under $\mathbb{F}_2-$arithmetic, I conjecture that $w$, the number of $1$'s in each row, should be an odd number. Consider a square matrix $n\times n$ matrix $A$, and the first element of one of its rows, say $a_{i1}$. I reason as follows:
a) Assume $w$ is an even number.
a1) Assume $a_{i1} = 0$. Then in the remaining $n-1$ elements of the row, there exists an even number of $1$'s, which by $\mathbb{F}_2-$arithmetic will give us $0$. The rest of the elements of the row all also zero, so overall the sum of the $n-1$ elements will give $0$, i.e equal to the value of the first element.
a2) Assume $a_{i1} = 1$. Then in the remaining $n-1$ elements of the row, there exists an odd number of $1$'s, which by $\mathbb{F}_2-$arithmetic will give us $1$. The rest of the elements of the row are all zero, so overall the sum of the $n-1$ elements will give $1$, i.e again equal to the value of the first element.
So if $w=even$, the $\mathbb{F}_2-$sum of the $n-1$ columns will always equal the $n$-th column, depriving us of full rank.
b) Assume $w$ is an odd number.
b1) Assume $a_{i1} = 0$. Then in the remaining $n-1$ elements of the row, there exists an odd number of $1$'s, which by $\mathbb{F}_2-$arithmetic will give us $1$. The rest of the elements of the row all zero, so overall the sum of the $n-1$ elements will give $1$, i.e different to the value of the first element.
b2) Assume $a_{i1} = 1$. Then in the remaining $n-1$ elements of the row, there exists an even number of $1$'s, which by $\mathbb{F}_2-$arithmetic will give us $0$. The rest of the elements of the row all also zero, so overall the sum of the $n-1$ elements will give $0$, i.e again different than the value of the first element.
So it appears that $w= odd$, is at least a necessary condition to have a matrix $A$ of full column/row rank.
4) Stochastic (in)dependence in the $\mathbb{F}_2-$world.
Does the characteristics of $\mathbb{F}_2-$field affect the concept of stochastic (in)dependence? No. Two r.v.'s are independent if and only if their joint distribution is the product of their marginals. Meaning, the conditional distributions must equal the unconditional ones. Maybe the way operations work in the $\mathbb{F}_2-$field produces some unexpected results?
Let's see: Assume that we have a square $A$ matrix that has $\mathbb{F}_2-$ linearly independent rows. The column vector process $\boldsymbol{v}'$, say of dimension $5\times 1$ is written
$$\boldsymbol{v}' = A\boldsymbol{v} = \left[\begin{matrix}
v_1'(v_0,...)\\
v_2'(v_0,...)\\
v_3'(v_0,...)\\
v_4'(v_0,...)\\
v_5'(v_0,...)\\
\end{matrix}\right]$$
Now assume that $w=3$ and, say, that the $1$'s in $A$ as dispersed such that we have
$$ v_2' = v_0+v_1+v_5,\qquad v_4' = v_2+v_3+v_5$$
Consider the conditional probability (under $\mathbb{F}_2-$ arithmetic)
$$P_{\mathbb{F}_2}(v_2' =1\mid v_4'=0) = P_{\mathbb{F}_2}(v_0+v_1+v_5 =1\mid v_2+v_3+v_5=0)$$
If the conditioning statement is to affect the probabilities of $v_2'$, it will do so through $v_5$: the fact that $v_2+v_3+v_5=0$ must affect the probabilities related to $v_5$. Under the "usual" arithmetic this would be obvious: it would mean that $v_5$ should equal zero. What happens under $\mathbb{F}_2-$ arithmetic?
We can examine
$$P_{\mathbb{F}_2}(v_5 =1\mid v_2+v_3+v_5=0) = \frac{P_{\mathbb{F}_2}(\{v_5 =1\}\land \{v_2+v_3+v_5=0\})}{P_{\mathbb{F}_2}(v_2+v_3+v_5=0)}$$
The possible values of $v_2+v_3+v_5$ under $\mathbb{F}_2-$ arithmetic are
from which we get the following contingency table
Therefore
$$P_{\mathbb{F}_2}(v_5 =1\mid v_2+v_3+v_5=0) = \frac{2p^2(1-p)}{(1-p)^3+3p^2(1-p)}=\frac{2p^2}{1-2p+4p^2} \neq p$$
except when $p=1/2$ - but the set up explicitly specifies that $p<1/2$ . So we conclude that
$$P_{\mathbb{F}_2}(v_5 =1\mid v_2+v_3+v_5=0) \neq P_{\mathbb{F}_2}(v_5 =1)$$
and so
$$P_{\mathbb{F}_2}(v_2' =1\mid v_4'=0) \neq P_{\mathbb{F}_2}(v_2' =1)$$
I have shown that, in general, the elements of the vector process $\boldsymbol{v}'$ are not independent (although they have identical marginals), even under $\mathbb{F}_2$-arithmetic. I have found and read the lemma in the paper the OP mentions. There is no actual proof, just a verbal assertion that they are independent, due to the $\mathbb{F}_2$-linear independence of the rows of $A$. Hopefully, I am mistaken, but for the moment it appears the assertion is not valid.
5) Now what?
Not much. The joint distribution of the random vector will depend on how the $1$'s are allocated in matrix $A$. Without a specific form, the various measures of distance between distributions become vacuous. Intuitively, if $w$ is small relative to the length of the rows of $A$, then one can expect/hope, that the dependence will be relatively weak, and so pretending that the vector is i.i.d. won't hurt much... but without knowing the joint distribution, one cannot really tell... Copulas for discrete r.v.'s suffer from identification issues... I am still thinking about it, but I am not optimistic.
Best Answer
Here I simply show the results of a sum of Bernoulli random variables where there is noise added to the probability parameter that follows a truncated Gaussian distribution, restricted to valid values of the parameter.
Let's assume we have a random variable $X$ which is, conditionally, a Bernoulli random variable.
$$X | \epsilon\sim\text{Bern}(p+\epsilon)$$ where $$\epsilon\sim TN(0,\sigma,a=0,b=1)$$ and $0<p<1$, $\sigma>0$ and $(a,b)$ represent the lower and upper truncation levels of the truncated Normal distribution, respectively. The truncated Normal distribution is used to ensure that the Bernoulli probability parameter is bounded between valid values.
We can simplify the above characterisation slightly as \begin{align} X | \epsilon\sim&\text{Bern}(\epsilon)\notag \end{align} where \begin{align} \epsilon\sim&TN(p,\sigma,a=0,b=1)\notag \end{align}
We can derive the probability mass function of $X$ using \begin{align} p_{X}(x)&=\int_{\epsilon}p_{X | \epsilon}(x | \epsilon)\, p_{\epsilon}(\epsilon)\,d\epsilon \notag \end{align} where \begin{align} p_{X | \epsilon}(x | \epsilon)&=\epsilon^{x}(1-\epsilon)^{1-x}\notag\\ &=\epsilon x+(1-\epsilon)(1-x)\notag \end{align} and \begin{align} p_{\epsilon}(\epsilon)&=\frac{e^{-\frac{1}{2}\big(\frac{\epsilon-p}{\sigma}\big)^{2}}}{\sqrt{2\pi}\sigma\big(\Phi(1;p,\sigma)-\Phi(0;p,\sigma)\big)}\notag\\ &=Ae^{-\frac{1}{2}\big(\frac{\epsilon-p}{\sigma}\big)^{2}}\notag \end{align}
So \begin{align} p_{X}(x)&=\int_{0}^{1}[\epsilon x+(1-\epsilon)(1-x)]Ae^{-\frac{1}{2}\big(\frac{\epsilon-p}{\sigma}\big)^{2}}\,d\epsilon \notag \end{align}
Let \begin{align} z=\epsilon-p\notag \end{align} Then \begin{align} dz=d\epsilon\notag \end{align}
\begin{align} p_{X}(x)&=\int_{-p}^{1-p}[(z+p)x+(1-z-p)(1-x)]Ae^{-\frac{1}{2}\big(\frac{z}{\sigma}\big)^{2}}\,dz \notag\\ &=\int_{-p}^{1-p}[px+(1-p)(1-x)]Ae^{-\frac{1}{2}\big(\frac{z}{\sigma}\big)^{2}} +(2x-1)zAe^{-\frac{1}{2}\big(\frac{z}{\sigma}\big)^{2}}\,dz \notag\\ &=px+(1-p)(1-x)+(2x-1)A\Big[-\sigma^{2}\Big(e^{-\frac{1}{2}\big(\frac{1-p}{\sigma}\big)^{2}}-e^{-\frac{1}{2}\big(\frac{p}{\sigma}\big)^{2}}\Big)\Big]\notag \end{align}
Which gives $$ p_{X}(x)= \left\{ \begin{aligned} (1-p) + \frac{\sigma\Big(e^{-\frac{1}{2}\big(\frac{1-p}{\sigma}\big)^{2}}-e^{-\frac{1}{2}\big(\frac{p}{\sigma}\big)^{2}}\Big)}{\sqrt{2\pi}\big(\Phi(1;p,\sigma)-\Phi(0;p,\sigma)\big)} &\quad\quad x=0\\ p - \frac{\sigma\Big(e^{-\frac{1}{2}\big(\frac{1-p}{\sigma}\big)^{2}}-e^{-\frac{1}{2}\big(\frac{p}{\sigma}\big)^{2}}\Big)}{\sqrt{2\pi}\big(\Phi(1;p,\sigma)-\Phi(0;p,\sigma)\big)} &\quad\quad x=1 \end{aligned} \right. $$
We can confirm this is a valid probability mass function by \begin{align} \sum_{x}p_{X}(x)&=p_{X}(x=0)+p_{X}(x=1)\notag\\ &=(1-p)+\frac{\sigma\Big(e^{-\frac{1}{2}\big(\frac{1-p}{\sigma}\big)^{2}}-e^{-\frac{1}{2}\big(\frac{p}{\sigma}\big)^{2}}\Big)}{\sqrt{2\pi}\big(\Phi(1;p,\sigma)-\Phi(0;p,\sigma)\big)}\notag\\ &\quad+p-\frac{\sigma\Big(e^{-\frac{1}{2}\big(\frac{1-p}{\sigma}\big)^{2}}-e^{-\frac{1}{2}\big(\frac{p}{\sigma}\big)^{2}}\Big)}{\sqrt{2\pi}\big(\Phi(1;p,\sigma)-\Phi(0;p,\sigma)\big)}\notag\\ &=1\notag \end{align}
Thus, we can say that \begin{align} X\sim\text{Bern}(p^{*})\notag \end{align} where \begin{align} p^{*}&=p-\frac{\sigma\Big(e^{-\frac{1}{2}\big(\frac{1-p}{\sigma}\big)^{2}}-e^{-\frac{1}{2}\big(\frac{p}{\sigma}\big)^{2}}\Big)}{\sqrt{2\pi}\big(\Phi(1;p,\sigma)-\Phi(0;p,\sigma)\big)}\notag \end{align}
We can see that \begin{align} \lim_{\sigma \to 0} \frac{\sigma\Big(e^{-\frac{1}{2}\big(\frac{1-p}{\sigma}\big)^{2}}-e^{-\frac{1}{2}\big(\frac{p}{\sigma}\big)^{2}}\Big)}{\sqrt{2\pi}\big(\Phi(1;p,\sigma)-\Phi(0;p,\sigma)\big)}=0\notag \end{align} which implies that as $\sigma \to 0$, the Bernoulli with Gaussian noise converges to a Bernoulli without noise, as expected. Furthermore, for $p=0.5$, the distribution of a Bernoulli with Gaussian noise is the same as that of a Bernoulli without noise.
Extension to the Binomial distribution is simple. Let's assume we have a sequence of $n$ independent Bernoulli trials with parameter $p^{*}$ and we define the sum as \begin{align} S_{n}&=\sum_{i=1}^{n}X_{i}\notag \end{align}
The moment generating function (MGF) of $S_{n}$ is \begin{align} M_{S_{n}}(t)&=(M_{X}(t))^n=(1-p^{*}+p^{*}e^{t})^{n}\notag \end{align}
which is the MGF of a $\text{Binomial}(n,p^{*})$ random variable.
Example We can illustrate the above with an example: $$p=\{0.3,0.7\},\, \sigma=0.5,\, n=20$$ where $n$ is the number of independent Bernoulli trials.
This gives $$p^{*}=\{0.442248,0.557752\}$$
The probability mass functions of the resultant Binomial random variables with and without noise are shown below.