I have deduced a proof as stated below and am not sure if it is correct, therefore need some advice.
Proof: Let $f(x) = ax^3+bx+c = 0$. $f(x)$ is continuous since it is a polynomial and it is differentiable since it has a limit. Assume $f(x)$ has $2$ roots, $f(a) = 0$ and $f(b) = 0$, there is a point $d\in(a,b)$ such that $f'(d) = 0$.
$$f'(x) = 3ax^2+b $$ Since $ab>0$, $a$ and $b$ must be both positive or both negative. $f'(d) = 3a(d)^2+b = 3ad^2+b$ not equal to $0$ instead $>0$ since any values of $d$ for $d^2$ will be positive.
Likewise $f'(d) = 3a(d)^2 + b$ will not equal to $0$ instead $<0$ for all negative values of $a$ and $b$.
Hence, a contradiction, $f$ has exactly one root.