It would help if you know the marginal of at least one variable $x_i$. Suppose this is the case, i.e. without loss of generality suppose that you know $p(x_1=0)$.
Then $p(x_1=i, x_2=j) = \left\{ \begin{array}{ll} p(x_1 = i) p(x_1=x_2) \quad &\mbox{if} \ i = j \\ p(x_1 = i) (1 - p(x_1 = x_2)) \quad & \mbox{if} \ i \neq j. \end{array} \right.$
Inductively, suppose we know $p(x_1 = i_1, ... , x_n = i_n)$. Then \begin{equation*} p(x_1 = i_1, ... x_n = i_n, x_{n+1} = i_{n+1}) = \left\{ \begin{array}{ll} p(x_1 = i_1,...,x_n=i_n) p(x_n=x_{n+1}) \quad &\mbox{if} \ i_n = i_{n+1} \\ p(x_1 = i_1,...,x_n=i_n) (1 - p(x_n=x_{n+1})) \quad & \mbox{if} \ i_n \neq i_{n+1}. \end{array} \right.\end{equation*} So this procedure gives the full distribution of $x_1,...,x_n$.
From this construction, you immediately notice that the family of distributions is fixed if you only have the marginal distribution of $x_1$ and all probabilities $p(x_n = x_{n+1})$. In particular, this raises the question what condition you require on your $p_{ij} = p(x_i = x_j)$, so that these are consistent.
This idea may be a start for a general solution, i.e. without the mentioned assumption.
Remark: there is an assumption of independence in my construction, which may not hold in your case. ($\{x_i=x_{i+1}\}$ is assumed to be independent of $x_1,...x_i$).