The idea is to introduce a new kind of coin, worth $0$ rubles and enforce that we use exactly $a-1$ coins and exactly $a$ coins respectively. Consider the formal series \[F_1(x, y) = (1+x^a+x^{2a}+\dots)\prod_{i = 0}^{b} (1+yx^i + (yx^i)^2 + (yx^i)^3 + \dots) = \frac{1}{1-x^a} \prod_{i=0}^{b} \frac{1}{1-yx^i}\]Observe that, for some nonnegative integer $n$ the coefficient of $x^n y^{a-1}$ in $F_1(x, y)$ counts the number of ways we can exchange an amount worth $n$ rubles into exactly $a-1$ coins and blue bills. Next, consider the formal series \[F_2(x, y) = (1+x^{a+b}+x^{2(a+b)} + \dots) \prod_{i = 0}^{b} (1+yx^i + (yx^i)^2 + (yx^i)^3 + \dots) = \frac{1}{1-x^{a+b}} \prod_{i=0}^{b} \frac{1}{1-yx^i}\]This time, the coefficient of $x^n y^a$ counts the number of ways we can exchange an amount worth $n$ rubles into exactly $a$ coins and red bills. The problem reduces to showing that for any nonnegative integer $n$ the coefficient of $x^n y^a$ in $yF_1(x, y)$ and $F_2(x, y)$ is the same, or that $y^a$ has the same coefficient in both $yF_1$ and $F_2$. Define $G(x, y) = yF_1(x, y)-F_2(x, y)$. We need to show that the coefficient of $y^a$ in $G(x, y)$ is null. Observe that we can safely replace $G$ with $G(1-x^a)(1-x^{a+b})$ because multiplying with polynomials in $x$ does not affect the coefficient of $y^a$ being zero. We have $G(x,y) = (y(1-x^{a+b})-(1-x^a))\prod_{i=0}^{b} \frac{1}{1-yx^i} = (x^a(1-yx^b) - (1-y))\prod_{i=0}^{b} \frac{1}{1-yx^i} = x^a \prod_{i=0}^{b-1} \frac{1}{1-yx^i} - \prod_{i=1}^{b} \frac{1}{1-yx^i}$. Now the end is near. Consider $f_1(x, y) = x^a \prod_{i=0}^{b-1} \frac{1}{1-yx^i}$ and $f_2(x, y) = \prod_{i=1}^b \frac{1}{1-yx^i}$. It suffices to show that for any nonnegative $n$ the coefficients of $x^n y^a$ in both $f_1$ and $f_2$ are identical. The coefficient of $x^n y^a$ in $f_1$ is the number of nonnegative integer sequences $u_i$ (for $1 \leq i \leq b$) with $a + \sum (i-1)u_i = n$ and $\sum u_i = a$ whereas the coefficient of $x^n y^a$ in $f_2$ is the number of nonnegative integer sequences $v_i$ (for $1 \leq i \leq b$) with $\sum iv_i = n$ and $\sum v_i = a$. It is obvious that any sequence satisfying the first constraint also satisfies the second, and vice versa, so these counts are identical and we have solved the original problem!