Step 1: Find the centre and radius of the first circle.
The circle $x^2 + y^2 + 2ax + c = 0$ can be rewritten as $(x+a)^2 + y^2 = a^2 - c$. So centre $C_1 = (-a, 0)$ and radius $r_1 = \sqrt{a^2 - c}$ (requires $a^2 > c$).
Step 2: Find the centre and radius of the second circle.
The circle $x^2 + y^2 + 2bx + c = 0$ gives $(x+b)^2 + y^2 = b^2 - c$. So centre $C_2 = (-b, 0)$ and radius $r_2 = \sqrt{b^2 - c}$ (requires $b^2 > c$).
Step 3: State the condition for one circle to lie completely inside another.
The first circle lies completely inside the second when the distance between the centres plus the radius of the inner circle is less than the radius of the outer circle: $|C_1 C_2| + r_1 < r_2$, i.e. $|{-a} - (-b)| + r_1 < r_2$, giving $|b - a| + \sqrt{a^2 - c} < \sqrt{b^2 - c}$.
Step 4: Deduce the size relationship of radii.
From $r_1 < r_2$ (since the first is inside the second): $a^2 - c < b^2 - c$, so $a^2 < b^2$, i.e. $|a| < |b|$.
Step 5: Deduce the sign relationship.
Both centres lie on the $x$-axis at $-a$ and $-b$ respectively. For the smaller circle to be completely inside the larger one, both centres must lie on the same side (otherwise the circles would straddle the origin and overlap rather than one containing the other). This means $-a$ and $-b$ have the same sign, i.e. $a$ and $b$ have the same sign, which means $ab > 0$.
Step 6: Verify that $ab > 0$ is the distinguishing condition.
If $ab < 0$, one centre is positive and the other negative, so the circles are on opposite sides of the $y$-axis and cannot be nested. The condition $c < 0$ ensures real radii but is not the classifying criterion among the given options.
Step 7: State the final answer.
\[ \boxed{ab > 0} \]