By multiplying $1+bc+cd+da$ and $1+ab$ together we get
$$(1+bc+cd+da)(1+ab) = 1+bc+cd+da+ab+ab^2c+abcd+a^2bd = \\ = 2+ab+bc+cd+da+\frac{b}{d}+\frac{a}{c}$$
From the inequality between the arithmetic and the geometric mean for the positive numbers $\frac{b}{d}$ and $\frac{a}{c}$, and from the equality $abcd=1$ we get $\frac{b}{d} + \frac{a}{c} \ge 2\sqrt{\frac{ab}{cd}} = 2ab$, and hence it holds that $(1+bc+cd+da)(1+ab) \ge 2+2ab+ab+bc+cd+da$
i.e.
$$1+bc+cd+da \ge 2+\frac{ab+bc+cd+da}{1+ab}$$
or
$$\frac{1+ab}{ab+bc+cd+da} \ge \frac{1}{bc+cd+da-1} \quad (1)$$
Analogously we get
$$\frac{1+bc}{ab+bc+cd+da} \ge \frac{1}{ab+cd+da-1} \quad (2)$$
$$\frac{1+da}{ab+bc+cd+da} \ge \frac{1}{ab+bc+cd-1} \quad (3)$$
and
$$\frac{1+da}{ab+bc+cd+da} \ge \frac{1}{ab+bc+cd-1} \quad (4)$$
By adding (1), (2), (3) and (4) together we get the inequality
$$\frac{4+ab+bc+cd+da}{ab+bc+cd+da} \geq \frac{1}{bc+cd+da-1} + \frac{1}{ab+cd+da-1} + \frac{1}{ab+bc+da-1} + \frac{1}{ab+bc+cd-1}$$
Since $ab + bc + cd + da \geq 4\sqrt[4]{(abcd)^2} = 4$ it follows that
$$\frac{1}{bc+cd+da-1} + \frac{1}{ab+cd+da-1} + \frac{1}{ab+bc+da-1} + \frac{1}{ab+bc+cd-1} \le 1 + \frac{4}{ab+bc+cd+da} \le 2$$
which was to be proven.

We will introduce the substitutions $ab = p$ and $bc = q$. According to that, $cd = \frac{1}{p}$ and $ad = \frac{1}{q}$, so the given inequality is equivalent to the inequality
$$\frac{1}{q + \frac{1}{p} + \frac{1}{q} - 1} + \frac{1}{p + \frac{1}{p} + \frac{1}{q} - 1} + \frac{1}{p + q + \frac{1}{q} - 1} + \frac{1}{p + q + \frac{1}{p} - 1} \le 2$$
On the other hand, if we use the inequalities $p + \frac{1}{p} \ge 2$ and $q + \frac{1}{q} \ge 2$ we get
$$\begin{aligned} & \frac{1}{q + \frac{1}{p} + \frac{1}{q} - 1} + \frac{1}{p + \frac{1}{p} + \frac{1}{q} - 1} + \frac{1}{p + q + \frac{1}{q} - 1} + \frac{1}{p + q + \frac{1}{p} - 1} \\ & \le \frac{1}{1+\frac{1}{p}} + \frac{1}{1+\frac{1}{q}} + \frac{1}{1+p} + \frac{1}{1+q} = \\ & = \frac{p}{p+1} + \frac{1}{p+1} + \frac{q}{q+1} + \frac{1}{q+1} = 2. \end{aligned}$$