Given: $x = 1 + a + a^2 + \ldots$ and $y = 1 + b + b^2 + \ldots$, where $|a| < 1$ and $|b| < 1$.

Since $|a| < 1$, the series $x$ is an infinite geometric progression with first term 1 and common ratio $a$. Using the formula for sum to infinity of a GP:

$x = \frac{1}{1-a}$

Similarly, since $|b| < 1$, the series $y$ is also an infinite GP with first term 1 and common ratio $b$:

$y = \frac{1}{1-b}$

Now, consider the series $1 + ab + a^2b^2 + \ldots$. This is a GP with first term 1 and common ratio $ab$. Since $|a| < 1$ and $|b| < 1$, we have $|ab| < 1$, so the sum to infinity exists:

$1 + ab + a^2b^2 + \ldots = \frac{1}{1-ab}$

Now, we compute the right-hand side of the required expression:

$xy = \frac{1}{(1-a)(1-b)}$

$x + y - 1 = \frac{1}{1-a} + \frac{1}{1-b} - 1$

$= \frac{(1-b) + (1-a) - (1-a)(1-b)}{(1-a)(1-b)}$

$= \frac{2 - a - b - 1 + a + b - ab}{(1-a)(1-b)}$

$= \frac{1 - ab}{(1-a)(1-b)}$

Therefore:

$\frac{xy}{x+y-1} = \frac{\dfrac{1}{(1-a)(1-b)}}{\dfrac{1-ab}{(1-a)(1-b)}}$

$= \frac{1}{1-ab}$

Hence, $1 + ab + a^2b^2 + \ldots = \dfrac{xy}{x+y-1}$, which proves the required result.

∴ $1 + ab + a^2b^2 + \ldots = \dfrac{xy}{x+y-1}$