Using pairings to verify an extended euclidean relation without leaking the values? Let $P_i(x)$ be polynomials $i=1,...,n$, $s$ some value, and $g$ a generator of a group $G$ where the discrete logarithm is hard.

Assume a prover wants to convince a verifier having access to the values $g^{P_i(s)}$ that it knows polynomials $q_i(x)$ such that the following equation holds:

$q_1(s)*P_1(s)+q_2(s)*P_2(s)+\ldots+q_n(s)*P_n(s) = 1.$ 

The prover thus sends the verifier the values $g^{q_i(s)}$ and
uses bilinear maps to verify the correctness of the answer.

Can the following equation tell me whether it has correctly computed the $q_i(s)$:
$$e(g^{q_i(s)},g^{P_i(s)}) = e (g,g)$$
where $e$ is a bilinear map $G\times G\rightarrow G_1$. 

So basically I want my bilinear map to verify on the exponents while hiding them. Is the second part of the equation correct? Or should it be 1, or $e(g,g)^{ord(G)}=1$?
Well, assuming the equation holds, $\Pi_{i=1}^n e(g^{q_i(s)},g^{P_i(s)}) = e (g,g)$ must also hold, due to the bilinearity of $e$. (Conversely, the equation only holds mod |G_1|.)

To see why, recall that the fact that the mapping $e$ is bilinear translates into $e(g^a,h^b)=e(g,h)^{a*b}$ for all elements $g$ and $h$ in $G$ and for all integers $a$ and $b$. Thus, we always have $e(g^{q_i(s)},g^{P_i(s)})=e(g,g)^{q_i(s)*P_i(s)}$ and by multiplying all these equalities together we get that $$\Pi_{i=1}^n e(g^{q_i(s)},g^{P_i(s)}) =e(g,g)^{q_1(s)*P_1(s)+\cdots+q_n(s)*P_n(s)}.$$
Now what you want to check is if $q_1(s)*P_1(s)+\cdots+q_n(s)*P_n(s)=1$: if it is true, we can replace the expression in the left-hand side of this equation in the pairing equation above by 1 and thus get
$$\Pi_{i=1}^n e(g^{q_i(s)},g^{P_i(s)}) = e (g,g)$$
as stated in the beginning.

For $e(g,g)$ is a group element of $G_1$, it must hold that $e(g,g)^{|G_1|}=1$, as for any other element of $G_1$.

For the actual value of $e(g,g)$ itself, we can only say it is an element of $G_1$ (by definition of $e$). It can be the neutral element of $G_1$ in which case $e(g^a,g^b)=1$ for all $a$ and $b$ and thus does not provide you with a way to check the extended euclidean equality, since you'll basically get $1=1$ for all possible values of $q_i$ and $P_i$.