A calculation by foot. We obtain from
\begin{align*}
\color{blue}{0}&\color{blue}{=\frac{1}{a+\omega}+\frac{1}{b+\omega}
+\frac{1}{c+\omega}+\frac{1}{d+\omega}-\frac{2}{\omega}}\tag{1}\\
\end{align*}
by multiplication with the common denominator $(a+\omega)(b+\omega)(c+\omega)(d+\omega)\omega$:
\begin{align*}
\color{blue}{0}&=(b+\omega)(c+\omega)(d+\omega)\omega\\
&\qquad+(a+\omega)(c+\omega)(d+\omega)\omega\\
&\qquad+(a+\omega)(b+\omega)(d+\omega)\omega\\
&\qquad+(a+\omega)(b+\omega)(c+\omega)\omega\\
&\qquad-2(a+\omega)(b+\omega)(c+\omega)(d+\omega)\\
&=bcd\omega+(bc+bd+cd)\omega^2+(b+c+d)\omega^3+\omega^4\\
&\qquad+acd\omega+(ac+ad+cd)\omega^2+(a+c+d)\omega^3+\omega^4\\
&\qquad+abd\omega+(ab+ad+bd)\omega^2+(a+b+d)\omega^3+\omega^4\\
&\qquad+abc\omega+(ab+ac+bc)\omega^2+(a+b+c)\omega^3+\omega^4\\
&\qquad-2abcd\\
&\qquad-2(abc+abd+acd+bcd)\omega\\
&\qquad-2(ab+ac+ad+bc+bd+cd)\omega^2\\
&\qquad-2(a+b+c+d)\omega^3\\
&\qquad-2\omega^4\\
&\,\,\color{blue}{=\left(2-(abc+abd+acd+bcd)\right)\omega-2abcd+a+b+c+d}\tag{2}
\end{align*}
In the last step (2) we observe that terms with $\omega^2$ cancel away and we also use the identities
\begin{align*}
\omega^3=1,\quad\omega^4=\omega
\end{align*}
We note from (1) and (2) we can write (1) as
\begin{align*}
\color{blue}{0=\frac{A\omega+B}{(a+\omega)(b+\omega)(c+\omega)(d+\omega)\omega}}\tag{3}
\end{align*}
and since $A,B\in\mathbb{R}$ and $\omega\in\mathbb{C}\setminus{\mathbb{R}}$ we conclude $A=B=0$, so that
\begin{align*}
\color{blue}{abc+abd+acd+bcd}&\color{blue}{=2}\tag{4}\\
\color{blue}{a+b+c+d}&\color{blue}{=2abcd}\\
\end{align*}
follows.
On the other hand we consider the expression
\begin{align*}
\color{blue}{\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}}&\color{blue}{=z}
\end{align*}
Multiplication of the LHS with the common denominator $(1+a)(1+b)(1+c)(1+c)$ gives
\begin{align*}
&(b+1)(c+1)(c+1)+(a+1)(c+1)(d+1)\\
&\qquad\quad+(a+1)(b+1)(d+1)+(a+1)(b+1)(c+1)\\
&\quad=1+(b+c+d)+(bc+bd+cd)+bcd\\
&\qquad\quad+1+(a+c+d)+(ac+ad+cd)+acd\\
&\qquad\quad+1+(a+b+d)+(ab+ad+bd)+abd\\
&\qquad\quad+1+(a+b+c)+(ab+ac+bc)+abc\\
&\quad=4+3(a+b+c+d)\\
&\qquad\quad+2(ab+ac+ad+bc+bd+cd)\\
&\qquad\quad+abc+abd+acd+bcd\\
&\,\,\color{blue}{=2\left(3+3abcd+ab+ac+ad+bc+bd+cd\right)}\tag{5}
\end{align*}
In the last line (5) we used the identities from (4).
Similarly, multiplication of the RHS with the common denominator gives
\begin{align*}
&z(a+1)(b+1)(c+1)(d+1)\\
&\qquad=z(1+(a+b+c+d)\\
&\qquad\quad+(ab+ac+ad+bc+bd+cd)\\
&\qquad\quad+(abc+abd+acd+bcd)+abcd)\\
&\,\,\color{blue}{\qquad=z(3+3abcd+ab+ac+ad+bc+bd+cd)}\tag{6}
\end{align*}
Again in the last line (6) we used the identities from (4) for simplification.
Comparing (5) and (6) we conclude
\begin{align*}
\color{blue}{\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}=2}
\end{align*}