Say I have some computation model/programming language $M$ (e.g. Turing machine or equivalent), and let $C_M$ be the set of all partial or total functions $f : \mathbb{N} \to \mathbb{N}$ computable by that model. Also, assume that $C_M$ is countable.
Now let $B = \{ B_1, B_2, ... , B_n \}$ a finite set of computable base functions and $\cdot : C_M \times C_M \to C_M$ an binary operator, which serves as a combination function. Thus, $(B,\cdot)$ is a magma.
I can now choose $B$ and $\cdot$ in such a way that I can decompose each computable function $f \in C_M$ into a finite product from $(B,\cdot)$, i.e. $\forall f \in C_M : \exists b_1, b_2, ..., b_k \in B : f = b_1 \cdot b_2 \cdot ... \cdot b_k$.
To show that, I consider that my programming language consists of a set of instructions $I$ and programs are words over $I$. Then, I choose exaclty one base function $b_{i,m}$ for each instruction $i_m$. Because each computable function $f$ has at least one program computing it, I can just pick one of those $f$-computing programs $P_f \in I^*$ and define $b_{i,a} \cdot b_{i,b} \cdot b_{i,c} ... = f$ when the program is the word $P_f = i_a i_b i_c ...$.
This decomposition is obviously not very useful, because it makes the base vectors arbitrary and the definition of $\cdot$ an infinite list of special cases. To consider it useful, the operator $\cdot$ should have some form of structure/finite definition, or in other words, be computable. My interest would be an element-wise function, so that given two values $f(x)$ and $g(x)$, it should be possible to compute $(f\cdot g)(x)$.
My question: Assuming $M$ is turing-complete, is it possible to choose $(B,\cdot)$ in such a way that $\cdot$ is elementwise computable, i.e.
$\exists h \in C_M : \forall f,g \in C_M : \forall x \in \mathbb{N} : (f\cdot g)(x) = h(f(x), g(x))$
and still have each computable function being decomposable into a finite product from $(B,\cdot)$?