Finite index subgroups of group extensions such that the quotient map is injective Suppose we have two groups $N$ and $G$. A group extension of $G$ by $N$ is a group $E$ that fits into the short exact sequence 
$0 \rightarrow N \rightarrow^i E \rightarrow^s  G \rightarrow 0.$
A nice example of an extension is given by semi-direct products, i.e., if we have a map $\phi : G \rightarrow \mbox{Aut}(N)$ then we construct semi-direct product $N \times_{\phi} G = \{(n,g) : n \in N, \ g \in G\}$ where $(n,g)(m,h) = (n \phi(g)(m),gh)$. (Note in this case $i(n) = (n,1)$ and $s(n,g) = g$.) If $\phi$ just maps to the identity automorphisms we recover the direct product of two groups. However, not all extensions are so nice.
My question is: 
If $N$ is a finite group (and $G = \langle S \vert R \rangle$ is finitely presented,) is there a subgroup $H \leq E$ of finite index such that $s$ restricted to $H$ is an isomorphism (or equivalently $H \cap i(N) = \{1\}$)? 
Why I think it might be true:
In the case of the semi-direct product, the answer is yes. Let $H = \{(1,g) : g \in G\}$, this is a subgroup of $N \times_{\phi} G$ using the multiplication I defined earlier, and $s\vert_{H}: H \rightarrow G$ is an isomorphisms.
What I have done so far:
Method 1:
Now consider a general extension. As $i(N) = \mbox{Ker}(s)$ it is normal in $E$. Therefore conjugation defines a map $\phi: E \rightarrow \mbox{Aut}(N)$ (this would be the same $\phi$ in the semi-direct case). Let $E_K = \mbox{Ker}(\phi)$ which is a finite index subgroup of $E$ as $N$ is finite therefore $\mbox{Aut}(N)$ is. Now define abelian group $N_K = i^{-1}(i(N) \cap E_K)$ (with $i(N_K) \leq Z(E_K)$ in the center of $E_K$), and group $G_K = s(E_K) \leq G$. Let $i_K$ and $s_K$ be $i$ and $s$ restricted to $A_K$ and $E_K$ respectively. One can show we get the following short exact sequence
$0 \rightarrow N_K \rightarrow^{i_K} E_K \rightarrow^{s_K}  G_K \rightarrow 0.$
this is what we call a central extension as $i(N_K) \leq Z(E_K)$. Central extensions are classified by maps (normalised 2-cocyles) $\psi: G_K \times G_K \rightarrow N_K$, where for $g_1,g_2,g_3,g \in G_K$
$\psi(g_1,g_2) \psi(g_1g_2,g_3) = \psi(g_1, g_2g_3) \psi(g_2,g_3)$, and
$\psi(g,g^{-1}) = \psi(g,1) = \psi(1,g) = 1$.
This classification dictates that $E_K = \{(n,g) : n \in N_K, \ g \in G_K\}$ where $(n,g)(m,h) = (nm \ \psi(g,h), gh)$. At this point I would like to use the two identies above to determine that there is another finite index subgroup $S \leq G_K$ such that $\psi(s_1,s_2) = 1$ for all $s_1,s_2 \in S$. This is where I am stuck.
Method 2:
Consider the set $S = \{H \leq E : H \cap i(N) = \{1\}\}$, this is clearly not empty as it contains the identity. By Zorn's lemma there is a maximal subgroup $M \in S$. If $M$ doesn't have finite index, there exists an element $g \in G \backslash (M \cup i(N))$ (else $G \backslash M = i(N) \backslash \{1\}$ which as $N$ is finite, is a finite set of elements bounding the order of index). The next step would be to pick a good element $g \in G \backslash (M \cup i(N))$ to add to $M$ to make a subgroup $M' = \langle g, M\rangle \geq M$ such that $M \cap N = \{1\}$, however, I am at a block here too.
 A: Lets start by restating your question (the re-statement allows us to ignore the map $i$, and also notes that $E$ is finitely presentable):

Question 1. If $N$ is a finite, normal subgroup of a finitely presentable group $E$, is there a subgroup $H \leq E$ of finite index such that $H \cap N = \{1\}$?

Okay, great. A group $K$ is residually finite if for every element $k\in K$ there exists a finite index normal subgroup $L_k$ of $K$ such that $k\not\in L_k$. This actually extends to finite sets: if $S$ is a finite subset of a finitely generated, residually finite group $K$ then there exists a finite index normal subgroup $L_S$ such that $S\cap L_S=\emptyset$ (take $L_S:=\cap_{k\in S}L_k$). As $N$ is a finite subgroup of $K$ we can take $H:=L_N$ to get the following:
Lemma. If $E$ is a finitely generated, residually finite group then the answer to the question is yes.
Great! But we can push this further. The finite residual of a group $K$ is the intersection of all finite index subgroups of $K$, so $\mathcal{R}:=\cap_{L\leq_fK}L$. This is a (normal) subgroup of $K$. We can rephrase your question in terms of the finite residual:

Question 2. If $N$ is a finite, normal subgroup of a finitely presentable group $E$, does $N$ intersect the finite residual $\mathcal{R}$ of $E$ trivially?

I think this view makes it clear that the answer to the question should be "no", as one would expect $\mathcal{R}$ to often contain torsion (although we would really need a finite subgroup of $\mathcal{R}$ which is characteristic in $\mathcal{R}$, which is stronger).
In the mean-time, we have an intermediate result.
Lemma. If $E$ is a finitely generated group whose finite residual is torsion-free then the answer to the question is yes.
Now lets answer the question: possibly there is an easy construction which will give an easy counter-example to the question, but instead lets use a ready-made construction of M. de Chiffre, L. Glebsky, A. Lubotzky and A. Thom, "Stability, cohomology vanishing, and nonapproximable groups." Forum of Mathematics, Sigma. Vol. 8. Cambridge University Press, 2020. (doi), where in Section 5.2 they prove:
Theorem. There are finitely presentable, residually finite groups $\Gamma$ which have finite central extensions
$$1\rightarrow C\rightarrow\widetilde{\Gamma}\rightarrow\Gamma\rightarrow1$$
where $C$ is a finite cyclic group and $\widetilde{\Gamma}$ is not residually finite.
As $\widetilde{\Gamma}/C$ is residually finite but $\widetilde{\Gamma}$ is not, it follows that $C$ contains the finite residual $\mathcal{R}$. Therefore, the finite residual is a finite cyclic subgroup of $\widetilde{\Gamma}$, and so we have our counter-example:
Corollary. The answer to the question is "no" in general.
A: If $G$ is finite as well, then a subgroup $H$, restricted to which $s$ becomes an isomorphism, is the trivial subgroup. But, as the cyclic group of order 4 shows (extension of $C_2$ by $C_2$), this might be the only choice. (This example generalizes in an obvious way to central products of such groups, but for the infinite product the quotient is not finitely generated, so I do not have an infinite example at hand.)
A: I think we get a counterexample with the nilpotent group $$E = \langle x,y,z \mid [x,z]=[y,z]=z^2=1,[x,y]=z \rangle$$ with $N=\langle z \rangle$  the central subgroup of order $2$ and $G = E/N \cong {\mathbb Z}^2$.
Then $N$ does not have a complement in $E$.
But after reading ahulpke's answer, I wondered whether I had misunderstood the question. I assumed you meant that $s$ restricted to $H$ should be an isomorphism onto $G$, but perhaps you just want $s$ to be injective?
