In this answer I will try to explain in details the proof of
Proposition 25 in Serre's book. For my convenience in writing this,
I will use Serre's notation so my $A$ is your $N$, i.e.
$G=H\ltimes A$ and $G_i=H_i\ltimes A$ where $H_i=\{h\in H:h\chi_i=\chi_i\}$
subgroup of $H$.
(Serre) The restriction of $\theta_{i,\rho}$ to $A$ only
involves characters $\chi$ belonging to the orbit $H\chi_i$
of $\chi_i$. This shows that $\theta_{i,\rho}$ determines $i$.
As the statement said, we consider character $\chi$ of $\theta_{i,\rho}$
for $a\in A$. Recall the character for induced representation for $s\in G$,
$H$ subgroup of $G$, $f:H\to \mathbf{C}^{\times}$, is
$$\text{Ind}_H^G(f)(s)=f'(s)=\frac{1}{|H|}\sum_{t\in G, t^{-1}st\in H}f(t^{-1}st)$$
Hence,
\begin{align*}
\chi(a) & = \frac{1}{|G_i|}\sum_{t\in G, t^{-1}at\in G_i}
\chi_{\chi_i\otimes \tilde{\rho}}(t^{-1}at), \\
& = \frac{1}{|G_i|}\sum_{t\in G} \chi_i(t^{-1}at)\chi_{\tilde{\rho}}(t^{-1}at),
\; (A \text{ normal so } t^{-1}at\in A\subset G_i \; \forall t\in G), \\
& = \frac{\text{dim }V}{|G_i|}\sum_{t\in G}\chi_i(t^{-1}at), \;
(t^{-1}at\in A \Rightarrow \tilde{\rho}(t^{-1}at)=\rho(1)=\text{id}_V), \\
& = \frac{\text{dim }V}{|G_i|} \sum_{a'\in A,h\in H}\chi_i((a'h)^{-1}a(a'h)),\\
& = \frac{\text{dim }V}{|G_i|} \sum_{a'\in A,h\in H} \chi_i(h^{-1}ah), \;
(A \text{ abelian}) \\
& = \frac{|A|\text{dim }V}{|G_i|} \sum_{h\in H} (h\chi_i)(a).
\end{align*}
Note that $h\chi_i\in X=\text{Hom}(A,\mathbf{C}^{\times})$ is
an irreducible representation of $A$ so the above implies that
the restriction of $\theta_{i,\rho}$ to $A$ is the direct sum of
representations corresponding to $h\chi_i$. Note that $h\chi_i$'s lie
in the orbit $H\chi_i\subset X$, which is disjoint with $H\chi_{i'}$
for $i\ne i'$. Hence, if $i\ne i'$, the restriction of $\theta_{i,\rho}$
and $\theta_{i',\rho'}$ to $A$ are not isomorphic.
(Serre) Let $W$ be the representation space for $\theta_{i,\rho}$, let
$W_i$ be the subspace of $W$ corresponding to $\chi_i$ [the set
of $x\in W$ such that $\theta_{i,\rho}(a)x=\chi_i(a)x$ for all
$a\in A$]. The subspace $W_i$ is stable under $H_i$ and the representation
of $H_i$ in $W_i$ is isomorphic to $\rho$; whence $\theta_{i,\rho}$
determines $\rho$.
Recall from $\S 7.1$ of Serre's book, $W$ can be identified with
$W=\mathbf{C}[G]\otimes_{\mathbf{C}[G_i]}V$ where
$V$ is the representation space of $\chi_i\otimes \tilde{\rho}$.
In particular, this is a $\mathbf{C}[G]$-module where $g$ acts
on $g'\otimes v$ by $(gg')\otimes v$. Also note that,
$(gg_i)\otimes v$ and $g\otimes (\chi_i\otimes \tilde{\rho})(g_i)v$
are considered the same in $W$ for $g_i\in G_i$.
Now, we identify $W_i$. For $a\in A, g\otimes v\in W$, we have
$\theta_{i,\rho}(a)(g\otimes v)= (ag)\otimes v
= g \otimes (g^{-1}ag)v$. Since $g^{-1}ag\in A$ so
$(\chi_i\otimes \tilde{\rho})(g^{-1}ag)v=\chi_i(g^{-1}ag)v$.
Thus, $\theta_{i,\rho}(a)(g\otimes v)= \chi_i(g^{-1}ag)
(g\otimes v)$. Thus, in order for
$\theta_{i,\rho}(a)(g\otimes v)=\chi_i(a)(g\otimes v)$ for all
$a\in A$, we must have $\chi_i(a)=(g\chi_i)(a)$ for all $a\in A$.
This follows $g\in H_i$. Thus, $W_i=\mathbf{C}[H_i]
\otimes_{\mathbf{C}[G_i]} V$.
With this, it is obvious that $W_i$ is stable under $H_i$.
Furthermore, observe $W_i$ is spanned by $1\otimes v_j$
where $v_j$'s basis of $V$, $1$ identity in $H_i$ (this holds
since $h\otimes v=1\otimes \rho(h)v$ for all $h\in H_i,v\in V$).
Hence, one can easily construct isomorphism representation of
$H_i$ in $W$ to $\rho$, as desired.
Best Answer
It's completely false. For example, the theory of quasisimple groups (groups $G$ such that $G$ is equal to its derived subgroup and $G/Z(G)$ is simple) comes under the theory of groups with an abelian normal subgroup. The theory of such groups is the fundamental block of the study of all characters of finite groups, called Clifford theory. If the extension is split then you can do what is in Serre, but in general you need to consider projective representations of the quotient $G$, which are homomorphisms from $G$ to $PGL_n$, not $GL_n$.
The faithful representations of, for example, $SL_n(q)$, which is such a quasisimple group, cannot be deduced from anything to do with $PSL_n(q)$, the quotient by the abelian normal subgroup.