Is there an identity for cos(ab)?

No, and there's a precise reason.

First, the geometric definition of $\cos$ talks about angles, and the product of two angles doesn't make sense.

Moreover, when you view the cosine as an exponential complex function, as you know $$\cos{x}= \frac{ e^{i x} + e^{-i x}}{2} $$ you can see that the identities you quoted come from properties of powers, such as $e^{a+b}=e^a e^b$ or $e^{2a} = (e^a)^2$

Since there's no significant formula for $e^{ab}$, there isn't one for the $\cos$ function too.

If $a$ is an integer and $b$ is an angle,

$$\cos(ab) = T_a(\cos b)$$

where $T_n(x)$ is the $n^{th}$ Chebyshev polynomial.

Not really, but I suppose this works: $$\cos ab=Re[(\cos(b)+i\sin(b))^a]$$

You can get the above equation by taking the real part of de Moivre's formula: $$\cos n\theta +i\sin n\theta=(\cos(\theta)+i\sin(\theta))^n \,$$

For general $a$ and $b$, we cannot write $\cos (ab)$ in terms of the trig functions $\cos a,\sin a, \cos b, \sin b$. This is because the trig functions are periodic with period $2\pi$, so adding $2\pi$ to $b$ does not change any of these functions. But adding $2\pi$ to $b$ can change $\cos (ab)$ - for instance, if $a=1/2$, if sends $\cos (ab)$ to $-\cos(ab)$. Only if $a$ is an integer can we avoid this problem.

As many experts already noted here, an argument of cos(⋯) is an angle, and a sensible mathematical structure on angles is the one of abelian-groups under “+”. We can add and subtract angles, as well as multiply them by integers. To some extent we can multiply angles by rational numbers, i.e. solve equations like $$ qx = pa,\quad x,a\text{ are angles, }\ p,q\in{\mathbb Z},\ q≠0.$$ If $a$ is specified modulo 2π radians, then such solutions, placed on the trigonometric circle, will form a set of $q$ elements (vertices of a regular $q$-gon), that can be described with an algebraic equation.

There is no “reasonable” multiplication of an angle and an irrational number $t$. If $a$ is specified modulo 2π radians (that is a typical condition), then possible values of $ta$ will form a dense subset of the trigonometric circle, and hence values of trigonometric functions on it will have no use for calculations.

Conclusion

There are trigonometric identities for products of an angle and a rational number; see other answers and this page for some partial cases.

There are no products of an angle and an irrational number, as well as there are no products of two angles.