Take \((a,b,c,d)=\left(\frac mn,\frac np,\frac pq,\frac qm\right)\). Then \(\sum\limits_{\rm cyc}\left(\frac mn+\frac nm\right)=0\), or \[(m+p)(n+q)\frac{mp+nq}{mnpq}=0.\]If $m+p=0$, then $ab=-1$; if $n+q=0$, then $ad=-1$; if $mp+nq=0$, then $ac=-1$.

You can also avoid the reparameterization as follows. We have $a+b+c+d+abc+abd+acd+bcd=0$, that is \[ a+b+c+d +ab(c+d) + cd(a+b) = (a+b)(1+cd)+(c+d)(1+ab) = 0, \]which together with $ab=1/cd$, yields \[ (1+cd)\left(a+b+\frac{c+d}{cd}\right) = 0 \Rightarrow (1+cd)\left(a(bc+1)+b(ad+1)\right) = 0, \]which once again using $ad=1/bc$ rearranges to \[ \frac1c(1+cd)(1+bc)(1+ac) = 0. \]From here, at least one of $ac,bc,cd$ is -1. From symmetry, the same holds also for $ab,ac,ad$ terms, concluding the proof.

We have that $a+b+c+d+abc+abd+acd+bcd=0$, multiplying by $a$ we get \[ a^2+ab+ac+ad+a^2bc+a^2bd+a^2cd+1 = 0 \Rightarrow (ab+1)(ac+1)(ad+1)=0 \]and we are done.

Let $ab=x^2$, $ac=y^2$, and $ad=z^2$. Because $abcd=1$ we get that $a=xyz$, $b=\frac{x}{yz}$, $c=\frac{y}{zx}$, and $d=\frac{z}{xy}$. Plugging into the second equality yields $$xyz+\frac{1}{xyz}+\frac{x}{yz}+\frac{yz}{x}+\frac{y}{zx}+\frac{zx}{y}+\frac{z}{xy}+\frac{xy}{z}=(x+\frac{1}{x})(y+\frac{1}{y})(z+\frac{1}{z})=0$$WLOG assume that $x+\frac{1}{x}=0$. Then we must have $x^2=ab=-1$, as desired.