This seems very obvious and I am having a bit of trouble producing a formal proof. sketch proof tha

Remark for your proof of composition of polynomials is a polynomial:
1. perhaps you should work with two arbitrary polynomials of degree $m$ and $n$ to have generality.
Remark for your proof of composition of composition rational functions is a rational function:
1. you wrote $a(z_1)=\frac{p(z_1)}{q(z_1)}$ and $b(z_2)=\frac{p(z_2)}{q(z_2)}$ which means $a$ and $b$ seems to be the same function.
2. ${\left(\frac{p(z_1)}{q(z_2)}\right)}^i$ is a rational function rather than a polynomial.
Guide for proof of composition of polynomials is a polynomial:
1. First prove that products of two polynomials is a polynomial. Once you can do that, we have that suppose $p$ is a polynomial, then $p(x{)}^i$ is a polynomials by mathematical induction.
2. Prove that the set of polynomials is closed under scalar multiplication.
3. Prove that the set of polynomials is closed under addition.
4. With those lemmas (tools), I believe now you can prove that composition of polynomials is a polynomials. (remember to use arbitrary polynomials of degree $m$ and degree $n$.)
Guide for proof of composition of rational functions is a rational function:
1. First prove that products of two rational function is a rational function. Once you can do that, we have that suppose $h$ is a rational function, then $h(x{)}^i$ is a rational function by mathematical induction.
2. Prove that the set of rational functions is closed under scalar multiplication.
3. Prove that the set of rational function is closed under addition.
4. Prove that the set of rational function is closed under division.
5. With those lemmas (tools), I believe now you can prove that composition of rational functions is a rational function. Let $a(z)=\frac{p(z)}{q(z)}$ and $b(z)=\frac{r(z)}{s(z)}$ and use those tools that you have verified.