It's right that the field of elliptic functions $K$ is generated by $\wp$ and$\wp^\prime$. To prove this, you consider the Laurent expansion of these two functions and use a comparison of coefficients and a variant of the Liouville theorem, which states that every elliptic function without a pole is constant. That wouldn't be possible if we used higher powers in the definition of $\wp$.
If we restrict ourselves to the one-dimensional case, there isn't anything like the Liouville theorem, since there are everywhere differentiable, bounded functions, who are nonconstant.
tl; dr: Yes.
Proposition: If $g$ is continuous and periodic, and if there is an integer $n > 1$ such that the function $h(x) = g(x^{n})$ is periodic, then $g$ is constant.
Here's a proof that (perhaps unfortunately) relies on concepts and results from real analysis: A continuous, periodic function on the reals is uniformly continuous on the reals.
Proof: It suffices to prove the contrapositive, i.e., if $g$ is continuous, periodic, and non-constant, and $n > 1$, then $h(x) = g(x^{n})$ is not periodic.
Assume $g$ is non-constant, let $\ell$ denote the smallest positive period of $g$, and consider the intervals $[(k\ell)^{1/n}, ((k+1)\ell)^{1/n}]$ for $k$ a positive integer, which map to the period intervals $[k\ell, (k+1)\ell]$ under $x \mapsto x^{n}$. Intuitively, the length of these intervals decreases to $0$ as $k$ grows without bound, but "$h$ goes through a complete period of $g$ on each", so $h$ is not periodic unless $g$ is constant.
Rigorously, if $g$ is not constant, there exist points $a$ and $b$ in $[0, \ell]$ such that $g(a) < g(b)$. It follows that for each positive integer $k$, there exist points $a_{k}$ and $b_{k}$ in $[(k\ell)^{1/n}, ((k+1)\ell)^{1/n}]$ such that $h(a_{k}) = g(a)$ and $h(b_{k}) = g(b)$. Since the difference $|h(b_{k}) - h(a_{k})|$ is independent of $k$ but $|b_{k} - a_{k}|$ decreases to $0$, $h$ is not uniformly continuous, and therefore not periodic.
Best Answer
All you need is for the ratio $\ \frac{a_1}{a_2}\ $ to be rational. If $\ \frac{a_1}{a_2}=\frac{k_1}{k_2}\ $, where $\ k_1,k_2\ $ are relatively prime integers, then $$ T=\frac{2\pi k_1}{a_1}=\frac{2\pi k_2}{a_2}\\ $$ is a period of both $\ \cos\big(a_1x+b_1\big)\ $ and $\ \cos\big(a_2x+b_2\big)\ $, and hence a period of $\ \cos\big(a_1x+b_1\big)+\cos\big(a_2x+b_2\big)\ $: \begin{align} \cos\big(a_1(x+nT)&+b_1\big)+\cos\big(a_2(x+nT)+b_2\big)\\ &=\cos\left(a_1\left(x+\frac{2\pi nk_1}{a_1}\right)+b_1\right)+\\ &\hspace{6em}\cos\left(a_2\left(x+\frac{2\pi nk_2}{a_2}\right)+b_2\right)\\ &=\cos\big(a_1x+b_1+2\pi nk_1\big)+\cos\big(a_2x+b_2+2\pi nk_2\big)\\ &=\cos\big(a_1x+b_1\big)+\cos\big(a_2x+b_2\big) \end{align}