Edit: I posted a wrong answer earlier. Hope I can get things fixed this time.
For convenience, write $x^T = (x_1^T, \ldots, x_m^T)$ where each $x_i$ is a vector of length $n$. Similarly, write $b^T = (b_1^T, \ldots, b_m^T)$. Let $J$ and $u$ be respectively the $n$-by-$n$ matrix and $n$-vector with all entries equal to 1 and let $Jb_i=\beta_iu$. Your system of equations $Ax=b$ is equivalent to $ (\dagger): (J-I)x_i+\sum_{j=1}^m x_j= b_i\quad \forall i. $ Let $\ Jx_i=\alpha_iu$ where $\alpha_i\in GF(2)$ is the parity of $x_i$. So $(\dagger)$ gives $ (*): x_i = \sum x_j + b_i + \alpha_iu\quad \forall i. $
Case 1: $m$ is even. By summing up $(*)$ from $i=1,2,\ldots,m$, we get $\sum x_j = \sum b_j + \sum\alpha_ju.$ Substitute this back into $(*)$, we see that the general solution to $(\ast)$ is of the form $ (**): x_i = \sum b_j + \sum\alpha_ju + b_i + \alpha_iu. $ Such $\{x_i\}$ form a solution of $(\dagger)$ if and only if $Jx_i=\alpha_iu$ for all $i$, which means $ \sum \beta_j + n\sum\alpha_j + \beta_i + n\alpha_i = \alpha_i $ or equivalently, $ n\sum\alpha_j + (n-1)\alpha_i = \sum \beta_j + \beta_i. $ When $n$ is even, the above system has a unique solution $\alpha_i = \sum \beta_j + \beta_i$.
When $n$ is odd, the above system reduces to $\sum\alpha_j = \sum \beta_j + \beta_i$. Hence solution exists if and only if $\beta_1=\ldots=\beta_m=\beta$ and the solutions are given by $(**)$ with $\sum\alpha_j = \beta$.
Case 2: $m$ is odd. By summing up $(*)$ from $i=1,2,\ldots,m$, we get $\sum b_j = \sum\alpha_ju.$ Thus a necessary condition for a solution to exist is that $\sum b_j$ is a multiple of $u$ (say, $\sum b_j=\lambda u$). If this is the case, then the general solution to $(\ast)$ is given by $x_i = y + b_i + \alpha_iu$ where $\sum \alpha_j=\lambda$ and $y$ is any $n$-vector. Such $\{x_i\}$ is a feasible solution to $(\dagger)$ if and only if $Jx_i=\alpha_iu$ for all $i$, that is, iff $Jy + \beta_iu + n\alpha_iu=\alpha_iu$. Therefore, we need $(n-1)\alpha_i+\beta_i$ to be constant and equal to the parity of $y$.
So, when $n$ is odd, solution exists only if $\beta_1=\ldots=\beta_m$. Since $ \sum b_j=\lambda u\ \Rightarrow\ \sum Jb_j=\lambda Ju\ \Rightarrow\ \sum\beta_j=\lambda, $ the previous requirement that $\sum \alpha_j=\lambda$ can be rewritten as $\sum \alpha_j=\sum \beta_j$.
When $n$ is even, that $(n-1)\alpha_i+\beta_i$ is constant means $\alpha_i+\beta_i=c$ for some constant $c$. Recall that we need $\sum \alpha_ju=\lambda u=\sum b_j$. Multiply both sides by $J$, we get $0=\sum \beta_ju$, i.e. $\sum \beta_j=0$. This is another necessary condition for the existence of solution. Suppose this is also satisfied. To make $\sum \alpha_ju=\lambda u$, we may take $\alpha_i=\beta_i$ for all $i$ if $\lambda=0$, or $\alpha_i=1-\beta_i$ for all $i$ if $\lambda=1$.