Omitting normalization is not a good idea in this case. The reason is because change in the normalization is a function of the state fed to the operator, so when you operate on a state that is not an eigenstate of the number operators then the balance among the states gets shifted.
The first step is to expand your generic state in the number operator basis:
$$|\psi\rangle = \sum_{\{n\}} \psi_{\{n\}} | n_1, n_2, \ldots n_k,\ldots\rangle .$$
We assume that this basis is, for our purposes, complete. If it is incomplete then that means that some of the number states are actually degenerate multiplets, but that's not relevant for this. Then the effect of raising and lowering operators should be obvious
\begin{align}
A(e_k)|\psi\rangle = \sum_{\{n\}} \sqrt{n_k} \psi_{\{n\}} | n_1, n_2, \ldots n_k - 1,\ldots\rangle \ \text{and}\\
A^\dagger(e_k)|\psi\rangle = \sum_{\{n\}} \sqrt{n_k+1} \psi_{\{n\}} | n_1, n_2, \ldots n_k + 1,\ldots\rangle.
\end{align}
Because $\psi$ isn't an eigenstate of the number operator, we cannot say that it has a definite number of particles that was increased by 1. We can, however, say that the expectation value of the number operator was increased by at least 1. The reason why the expected number can increase by more than one is because the raising operator also up-weights the high number states.
Here's some of the math behind that. First, we need to compute the normalization constant:
$$\mathcal{N}^2 = \langle \psi|A(e_k) A^\dagger(e_k)|\psi\rangle = \sum_{\{n\}} (n_k+1) \psi_{\{n\}}^* \psi_{\{n\}}.$$
Next, we insert a number operator $N_j = A^\dagger(e_j) A^\dagger(e_j)$ to get the expectation:
$$\langle \psi|A(e_k) A^\dagger(e_j) A(e_j) A^\dagger(e_k)|\psi\rangle = \mathcal{N}^{-1}\sum_{\{n\}} (n_k+1) (n_j + \delta_{jk}) \psi_{\{n\}}^* \psi_{\{n\}},$$
with $\delta_{jk}$ the Kronecker delta (it equals 1 if $j=k$, 0 otherwise).
Let's now consider a concrete example of a space with two excitation types. Specifically, let's assume that $\psi = \frac{1}{2} e^{-\ln(2)(n_1+n_2) / 2}$, so that each excited state is twice as common as the next higher state. First, let's compute the expected number of particles in each state:
\begin{align}
\langle n_i\rangle &= \sum_{n_i=0}^\infty n_i \left(\frac{1}{2}\right)^{n_i+1} \\
&= \left[\frac{1}{2}\frac{\partial}{\partial r} \sum_{n_i=0}^\infty r^{n_i}\right]_{r=1/2} \\
&= \left[\frac{1}{2}\frac{\partial}{\partial r} \frac{1}{1-r} \right]_{r=1/2} \\
&= 2
\end{align}
Next, our normalization constant when we raise the first number type:
\begin{align}
\mathcal{N} &= \frac{1}{2}\sqrt{\sum_{\{n\}} (n_k+1) \psi_{\{n\}}^* \psi_{\{n\}}} \\
&= \frac{1}{2}\sqrt{\sum_{n_1=0}^\infty\sum_{n_2=0}^\infty (n_1+1) \left(\frac{1}{2}\right)^{n_1+n_2}} \\
&= \frac{1}{2}\sqrt{\sum_{n_1=0}^\infty (n_1+1) \left(\frac{1}{2}\right)^{n_1}} 2 \\
&= \frac{1}{2}\sqrt{4 + \sum_{n_1=0}^\infty \left.r\frac{\partial r^{n_1}}{\partial r}\right|_{r=1/2} 2} \\
&= \frac{1}{2}\sqrt{4 + \left.r\frac{\partial}{\partial r}\left[\frac{1}{1-r}\right]\right|_{r=1/2} 2} \\
&=\sqrt{2}.
\end{align}
Next, the expected number of type 1 is
\begin{align}
\langle \psi|A(e_1) A^\dagger(e_1) A^\dagger(e_1) A^\dagger(e_1)|\psi\rangle &= \frac{1}{\sqrt{8}}\sum_{n_1=0}^\infty\sum_{n_2=0}^\infty (n_1+1)^2 \left(\frac{1}{2}\right)^{n_1+n_2} \\
&= \frac{1}{\sqrt{2}} \left[\frac{\partial}{\partial r} \left(r\frac{\partial}{\partial r}\right) \sum_{n_1=0}^\infty r^{n_1+1} \right]_{r=1/2} \\
&= \frac{12}{\sqrt{2}},
\end{align}
which is an increase greater than 1.
Because the states in our example factor, the $n_i$ are independent. If we had chosen a state where they don't factor, for example $\psi \propto \frac{1}{\cosh(2 \ln(2)(n_1+n_2))}$, $\frac{1}{(n_1 + n_2)^4}$, or $\left(n_1 + n_2^{-1}\right) \left(\frac{1}{2}\right)^{(n_1+n_2)/2}$, then the raising operator on type 1 would also affect the expected number of type 2. In fact, raising type 1 could raise or lower type 2, depending on how they correlate in the original state.
Lowering can have even more counter-intuitive results because it destroys the ground state. Consider $|\psi\rangle = \frac{4}{5}|0\rangle + \frac{3}{5}|10\rangle$. The expected number in the base state is $\langle N\rangle = \frac{18}{5}=3.6$. After applying the lowering operator and normalizing, our state becomes
\begin{align}
\frac{A |\psi\rangle}{\sqrt{\langle \psi|A^\dagger A|\psi\rangle}} &= |9\rangle.
\end{align}
In other words, the lowering operator increased the expected number for $\psi$ because it annihilated the contribution from the ground state, which had a higher weight in the original sate.
I leave a general proof that raising increases the expected single particle number by 1 or more, and whether anything can be said about the general effects of the lowering operator, as exercises for the reader.
