Solution 1. Note that
$$\begin{align*} (ad + bc + ca - db)^2 &= (a(c+d) + b(c-d))^2 \\ &= a^2(c+d)^2 + 2ab(c^2-d^2) + b^2(c-d)^2 \\ &\leq (c+d)^2 + 2ab(c^2-d^2) + (c-d)^2 \\ &= 2(c^2+d^2) + 2ab(c^2-d^2) \\ &= 2[(1+ab)c^2 + (1-ab)d^2] \\ &\leq 2[(1+ab) + (1-ab)] \quad (\text{since } -1 \leq ab \leq 1) \\ &= 4, \end{align*}$$
whence
$$|ad + bc + ca - db| \leq 2.$$

When $a^2 \neq 1$, equality can only occur when we are in case (iii), hence $b^2 = 1$.
Then $ab \neq \pm 1$ and equality in the second inequality occurs iff $c^2 = d^2 = 1$.
When $b^2 \neq 1$, equality can only occur when we are in case (ii), hence $a^2 = 1$.
Then $ab \neq \pm 1$ and equality in the second inequality occurs iff $c^2 = d^2 = 1$.
When $a^2 = b^2 = 1$, we either have $a = b = \pm 1$ and then $ab = 1$, in which case equality in the second inequality occurs iff $c^2 = 1$, or we have $a = -b = \pm 1$, hence $ab = -1$, and equality in the second inequality occurs iff $d^2 = 1$.

Equality holds in the first inequality iff we are in one of the following four cases: (i) $a^2 = b^2 = 1$, or (ii) $a^2 = 1, c = d$, or (iii) $b^2 = 1, c = -d$, or (iv) $c = d = 0$. When $c = d = 0$, the second inequality is strict, so we are not interested in case (iv).
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In summary, equality occurs iff one of the following holds:
$$\begin{aligned} a &= b \quad \text{and} \quad a^2 = b^2 = c^2 = 1, \text{ or} \\ c &= -d \quad \text{and} \quad b^2 = c^2 = d^2 = 1. \end{aligned}$$

Solution 2. Note that if $x$, $y$ are real numbers then
$$|x + y| + |x - y| = 2 \max(|x|, |y|).$$
For
$$\begin{aligned} (|x + y| + |x - y|)^2 &= (x + y)^2 + 2|x + y||x - y| + (x - y)^2 \\ &= 2(x^2 + y^2) + 2|x^2 - y^2| \\ &= 4 \max(x^2, y^2) \\ &= 4(\max(|x|, |y|))^2. \end{aligned}$$
Hence
$$\begin{aligned} |ad + bc + ca - db| &= |a(c+d) + b(c-d)| \\ &\leq |a| |c+d| + |b| |c-d| \\ &\leq |c+d| + |c-d| \\ &= 2 \max(|c|, |d|) \\ &\leq 2. \end{aligned}$$

The case of equality can be determined in a way similar to Solution 1.