Say we are working in $\mathbb{F}^{2n}_2$ where vectors can be written as pairs $(a,b)$ with $a,b \in \mathbb{F}^{n}_2$. Given a list of basis vectors for an $n-1$ dimensional subspace $S \subset \mathbb{F}^{2n}_2$, I want a polynomial time algorithm that can test membership of an input vector $(a,b)$.

This version of the problem is easy to solve using coding theory: I construct a generator matrix of the null space of $S$, call it $S^\perp$, and use $S^\perp$ as a parity check matrix: if $S^\perp(a,b) = \mathbf{0}$ then $(a,b) \in S$.

I am interested in a weird case where my vectors have an additional sign bit: $(s,a,b)$. Addition is defined as: $$(s,a,b) + (s',a',b') = (s + s' + a^Tb' + b^Ta', a+a', b+b')$$ How can I test for membership in this case? By ignoring the sign bit I can use the parity check test to determine membership of the $(a,b)$ part, but since this method does not tell me a decomposition into basis vectors I can't use the result to verify if the sign is correct. Let's say either $(0,a,b) \in S$ or $(1,a,b) \in S$, but never both.

(To those who want context, this is a question about $[[n,1]]$ quantum stabilizer codes: Given a list of stabilizer generators, test for membership of a given Pauli operator, taking the sign of the operator into account.)

For example, let the rows of the following matrix be a basis for $S$ when $n = 4$: $$\begin{bmatrix}1, & 1&0&0&1,&0&0&0&0\\ 0,&0&1&0&0,&0&1&1&0\\0,&0&0&1&0,&0&1&1&0 \end{bmatrix}$$ Is the vector $(0,1111,0000)$ in this space? The parity check method says yes, but in fact the vector has the wrong sign. How can I verify this without expanding out the whole space?