You have $r$ properties, denoted by $a_1,a_2,\dots,a_r$. For any set $S\subseteq\{a_1,\dots,a_r\}$ of properties, $N(S)$ is the number of objects having all of the properties in $S$. For example, $N\big(\{a_1\}\big)$ is the number of objects with property $a_1$, and $N\big(\{a_1,a_2\}\big)$ is the number with both property $a_1$ and property $a_2$. (This notation is slightly different from the notation in the question, but it makes things a little easier.) For $i=0,\dots,r$, $e_i$ is the number of objects with exactly $i$ of the $r$ properties, and $s_i$ is the number with at least $i$ of the properties. The total number of objects is $n$; clearly every object has at least $0$ of the properties, so $s_0=n$.
It’s easy to express the numbers $s_i$ in terms of the numbers $N(a_k)$:
$$s_i=\sum\left\{N(S):S\subseteq\{a_1,\dots,a_r\}\text{ and }|S|=i\right\}\;.$$
E.g., for $i=1$ this is just $$\sum_{k=1}^rN\big(\{a_k\}\big)\;.$$
We want to prove that for $m=0,\dots,r$,
$$\begin{align*}
e_k&=s_k-\binom{k+1}1s_{k+1}+\ldots+(-1)^r\binom{r}{r-m}s_r\\
&=\sum_{m=k}^r(-1)^{m-k}\binom{m}{m-k}s_m\\
&=\sum_{m=k}^r(-1)^{m-k}\binom{m}ks_m\;.
\end{align*}$$
There are several ways to do this, so the one that I’m going to use may not be the one that you’ve already seen.
Suppose that $x$ is an object with at least $m$ of the $r$ properties. Then there is a $k$ such that $m\le k\le r$ and $x$ has exactly $k$ of the properties. Let $P$ be the set of $k$ properties possessed by $x$; $P$ has $\binom{k}m$ subsets of size $m$, and $x$ is counted once for each of them in the calculation of $e_m$, so $x$ contributes a total of $\binom{k}m$ to $e_m$. It follows that the $e_k$ objects with $k$ of the properties contribute a total of $\binom{k}me_k$ to $e_m$ and hence that $$s_m=\sum_{k=m}^r\binom{k}me_k\;.\tag{1}$$
Now I’ll define two polynomials: let
$$S(x)=\sum_{k=0}^rs_kx^k\quad\text{and}\quad E(x)=\sum_{k=0}^re_kx^k\;.$$
In view of $(1)$ we have
$$\begin{align*}
S(x)&=\sum_{m=0}^rs_mx^m\\
&=\sum_{m=0}^r\sum_{k=m}^r\binom{k}me_kx^m\\
&=\sum_{k=0}^r\sum_{m=0}^k\binom{k}me_kx^m&&\text{reverse the order of summation}\\
&=\sum_{k=0}^re_k\sum_{m=0}^k\binom{k}mx^m\\
&=\sum_{k=0}^re_k(x+1)^k&&\text{by the binomial theorem}\\
&=E(x+1)\;.
\end{align*}$$
Of course this implies that $E(x)=S(x-1)$, and we can reverse the above calculation to find that
$$\begin{align*}
E(x)&=S(x-1)\\
&=\sum_{m=0}^rs_m(x-1)^m\\
&=\sum_{m=0}^rs_m\sum_{k=0}^m(-1)^{m-k}\binom{m}kx^k&&\text{binomial theorem again}\\
&=\sum_{k=0}^r\sum_{m=k}^r(-1)^{m-k}\binom{m}ks_mx^k&&\text{reverse the order again}\;.
\end{align*}$$
The coefficient of $x^k$ in $E(x)$ is therefore
$$e_k=\sum_{m=k}^r(-1)^{m-k}\binom{m}ks_m\;,$$
exactly as we wanted.
Here we follow closely the proof of Exercise 3, chapter 2: Sieve methods from Enumerative Combinatrics, Vol. I by R. P. Stanley. To ease comparison I'll try to use OP's notation where appropriate.
Setting: We consider a set $A=\{P_1,\ldots,P_n\}$ of $n$ properties. Given a finite set $X$ we denote
with $E_{m}$ the number of elements in $X$ where each element has exactly $m$ properties from $A$.
with $E_{\ge m}$ the number of elements in $X$ where each element has at least $m$ properties from $A$.
The claim: The following is valid
\begin{align*}
\color{blue}{E_m=\sum_{j=m}^n(-1)^{j-m}\binom{j}{m}S_j}\tag{1}
\end{align*}
given
\begin{align*}
\color{blue}{S_j=\sum_{{T\subseteq A}\atop{|T|=j}}E_{\geq}(T)}\tag{2}
\end{align*}
where $E_{\geq}(T)$ means the number of elements in $X$ where each element has at least the properties from $T\subseteq A$.
Proof: We start with the right-hand side from (1) and put (2) into it. We so obtain
\begin{align*}
\color{blue}{\sum_{j=m}^n}&\color{blue}{(-1)^{j-m}\binom{j}{m}S_j}\\
&=\sum_{j=m}^n(-1)^{j-m}\binom{j}{m}\sum_{{T\subseteq A}\atop{|T|=j}}E_{\geq}(T)\tag{3.1}\\
&=\sum_{j=m}^n(-1)^{j-m}\binom{j}{m}
\sum_{{T\subseteq R\subseteq A}\atop{|T|=j}}E_{=}(R)\tag{3.2}\\
&=\sum_{R\subseteq A}E_{=}(R)\sum_{T\subseteq R}(-1)^{|T|-m}\binom{|T|}{m}\tag{3.3}\\
\end{align*}
Comment:
Now we look at the inner sum of (3.3) which can be considerably simplified. We obtain setting $|R|=r$
\begin{align*}
\color{blue}{\sum_{T\subseteq R}(-1)^{|T|-m}\binom{|T|}{m}}&
=\sum_{j=0}^r\sum_{{T\subseteq R}\atop{|T|=j}}(-1)^{|T|-m}\binom{|T|}{m}\tag{3.4}\\
&=\sum_{j=0}^r(-1)^{j-m}\binom{j}{m}\sum_{{T\subseteq R}\atop{|T|=j}}1\tag{3.5}\\
&=\sum_{j=0}^r(-1)^{j-m}\binom{j}{m}\binom{r}{j}\tag{3.6}\\
&=\binom{r}{m}\sum_{j=0}^r(-1)^{j-m}\binom{r-m}{r-j}\tag{3.7}\\
&\,\,\color{blue}{=\delta_{r,m}}\tag{3.8}
\end{align*}
Comment:
In (3.4) we rearrange the sum according to the size $j$ of $T$, $0\leq j\leq r$.
In (3.5) we factor out the sign and the binomial coefficient leaving a summand $1$.
In (3.6) we note that $\binom{r}{j}$ is the number of subsets with $j$ elements of a set with size $r$.
In (3.7) we use the binomial identity $\binom{j}{m}\binom{r}{j}=\binom{r}{m}\binom{r-m}{r-j}$.
In (3.8) we use $\sum_{j=0}^{r}(-1)^j\binom{r-m}{r-j}=(1-1)^{r-m}=\delta_{r,m}$.
Putting finally (3.8) into (3.3) we obtain
\begin{align*}
\color{blue}{\sum_{j=m}^n(-1)^{j-m}\binom{j}{m}S_j}
&=\sum_{R\subseteq A}E_{=}(R)\delta_{|R|,m}\\
&=\sum_{{R\subseteq A}\atop{|R|=m}}E_{=}(R)\\
&\,\,\color{blue}{=E_{m}}
\end{align*}
and the claim follows.
Addendum: The proof revisited
Here we look at the two key ingredients which R. P. Stanley used in the proof.
We start with a comparison of the claim (1) and the binomial expression in (3.6)
\begin{align*}
E_m=&\sum_{j=m}^n\color{blue}{(-1)^{j-m}\binom{j}{m}}S_j\tag{1}\\
&\sum_{j=m}^n\color{blue}{(-1)^{j-m}\binom{j}{m}}\binom{r}{j}=\delta_{r,m}\tag{3.6, 3.8}\\
\end{align*}
It is the binomial identity (3.6,3.8) which is the reason for choosing the coefficient $(-1)^{j-m}\binom{j}{m}$ in (1). Since this cancels all unwanted terms leaving finally those of $E_m$.
The second ingredient is the rearrangement of $S_j$ from a representation with $E_{\geq}(\cdot)$ summands into a representation with $E_{=}(\cdot)$ summands. We have in steps (3.3) to (3.8) the following rearrangement of $S_j$:
\begin{align*}
S_j=\sum_{{T\subseteq A}\atop{\color{blue}{|T|=j}}}\color{blue}{E_{\geq}(T)}
&=\sum_{{T\subseteq R\subseteq A}\atop{|T|=j}}E_{=}(R)\\
&=\sum_{R\subseteq A}E_{=}(R)\sum_{{T\subseteq R}\atop{|T|=j}}1\\
&\,\,=\sum_{R\subseteq A}\color{blue}{E_{=}(R)\binom{|R|}{j}}\tag{3.9}
\end{align*}
It is this representation (3.9) which admits the usage of the binomial identity from above, since putting (3.9) into (1) we obtain
\begin{align*}
\color{blue}{\sum_{j=m}^n(-1)^{j-m}\binom{j}{m}S_j}&
=\sum_{R\subseteq A}E_{=}(R)\sum_{j=m}^n(-1)^{j-m}\binom{j}{m}\binom{|R|}{j}\\
&=\sum_{R\subseteq A}E_{=}(R)\,\delta_{|R|,m}\\
&=\sum_{{R\subseteq A}\atop{|R|=m}}E_{=}(R)\\
&\,\,\color{blue}{=E_m}
\end{align*}
which completes the proof.
Best Answer
This is a rather pragmatic, semi-combinatorial answer. Nevertheless it might be useful for some more creative combinatorialists. We provide an interpretation of the binomial coefficients of $E_m$ and derive from them an interpretation of the binomial coefficients of $L_m$.
At first I'd like to introduce a slightly different notation.
Interpretation of binomial coefficients of $E_{=m}$:
We can now generalise:
Interpretation of binomial coefficients of $L_{\geq m}$:
Note: In order to see some more aspects regarding $\binom{j}{m}$, the coefficient of $S_j$ in $E_m$, we give here a proof of (1) following (2.39) in Richard P. Stanleys Enumerative Combinatorics, Vol. 1, Ed.02.
We obtain \begin{align*} \color{blue}{\sum_{j=m}^n}&\color{blue}{(-1)^{j-m}\binom{j}{m}S_j}\\ &=\sum_{j=m}^n(-1)^{j-m}\binom{j}{m}\sum_{{Q\subseteq P}\atop{|Q|=j}}L_{\geq} (Q)\\ &=\sum_{j=m}^n(-1)^{j-m}\binom{j}{m}\sum_{{Q\subseteq R\subseteq P}\atop{|Q|=j}}E_{=}(R)\\ &=\sum_{R\subseteq P}E_{=}(R)\sum_{Q \subseteq R}(-1)^{|Q|-m}\binom{|Q|}{m}\\ &=\sum_{R\subseteq P}E_{=}(R)\sum_{j=0}^{|R|}(-1)^{j-m}\binom{|R|}{j}\binom{j}{m}\\ &=\sum_{R\subseteq P}E_{=}(R)\binom{|R|}{m}\sum_{j=0}^{|R|}(-1)^{j-m}\binom{|R|-m}{|R|-j}\\ &=\sum_{R\subseteq P}E_{=}(R)\binom{|R|}{m}\delta_{|R|,m}\\ &\,\,\color{blue}{=E_{=m}} \end{align*} showing the validity of (1).