Is there any way to solve for $a$ and $b$ in:
$$ a*b \equiv s_0 \mod r_0 $$
$$ a - b - a*b \equiv s_1 \mod ( r_0 - 1) $$
I have the values of $r_0$, $s_0$, $s_1$ and would like to find the values for $a$, $b$? The $*$ is just normal integer multiplication.
Write the second equation as $$(a+1)(b-1) = ab + b - a - 1 \equiv -s_1 - 1 \mod (r_0 - 1) $$ Take any $x,y,z,w$ such that
$$x y \equiv s_0 \mod r_0$$
$$z w \equiv -s_1 - 1 \mod (r_0-1)$$.
Since $r_0$ and $r_0-1$ are coprime, we can find $a$ and $b$ such that:
$$a \equiv x \mod r_0$$
$$b \equiv y \mod r_0$$
$$a \equiv z-1 \mod (r_0 - 1)$$
$$b \equiv w+1 \mod (r_0 - 1)$$.
Then these satisfy your equations.
EDIT: For example, if $r_0 = 5$, $s_0 = 1$ and $s_1 = 2$, you want $x y \equiv 1 \mod 5$ and $zw \equiv -3 \equiv 1 \mod 4$. One possibility (there are many) is $x \equiv 2 \mod 5$, $y \equiv 3 \mod 5$, $z \equiv 3 \mod 4$, $w \equiv 3 \mod 4$. Then using the Chinese Remainder Theorem, $a \equiv 2 \mod 20$ and $b \equiv 8 \mod 20$. And indeed $ab \equiv 2 \cdot 8 \equiv 1 \mod 5$ and $a-b-ab \equiv 2 - 8 -2\cdot 8 \mod 4$.