Actually, these two conjectures are actually equivalent.
We need a lemma: Lucas' Theorem: $\binom{a}{b}$ is odd if and only if $a\&b=b$ where $\&$ is bitwise AND operation. Let $S_j$ be the set of integers $i$ such that $i\&j=i$. Thus, if the first relation
$$a(n, -1) = \sum\limits_{j=0}^{n}(-1)^{\operatorname{wt}(n)-\operatorname{wt}(j)}(\binom{n}{j}\operatorname{mod} 2)a(j,0)$$ holds, we have $$a(n, -1) = \sum\limits_{j\in S_n}(-1)^{\operatorname{wt}(n)-\operatorname{wt}(j)}a(j,0)$$ And therefore, $$\sum_{j=0}^{n}(\binom{n}{j}\operatorname{mod} 2)a(j,-1)=\sum_{j\in S_n}a(j,-1)=\sum_{j\in S_n}\sum_{k\in S_j}(-1)^{\text{wt}(j)-\text{wt}(k)}a(k,-1)$$ Swapping the sums, we have the right hand side equals to $$\sum_{k\in S_n}a(k,-1)\sum_{j\text{ such that }k\in S_j,j\in S_n}(-1)^{\text{wt}(j)-\text{wt}(k)}$$ All the other $k\in S_n$ vanishs the sum $\sum_{j\text{ such that }k\in S_j,j\in S_n}(-1)^{\text{wt}(j)-\text{wt}(k)}$ because for every bit that is $1$ in $n$ and $0$ in $k$, it contribute a factor $1$ if that bit in $j$ is $0$ and $-1$ if that bit in $j$ is $1$. Therefore, the sum $$\sum_{k\in S_n}a(k,-1)\sum_{j\text{ such that }k\in S_j,j\in S_n}(-1)^{\text{wt}(j)-\text{wt}(k)}$$ is actually $a(n,-1)$, which implies the conjecture $a(n, 0) = \sum\limits_{j=0}^{n}(\binom{n}{j}\operatorname{mod} 2)a(j,-1)$. Similarly, we can derive the first conjecture using the second.
So, we only need to proof $a(n, 0) = \sum\limits_{j=0}^{n}(\binom{n}{j}\operatorname{mod} 2)a(j,-1)=\sum_{j\in S_n}a(j,-1)$.
(I have proved, but I need some time to write)