Let $$d(x,y)= \begin{cases} 1, &\text{if }x\text{ is divisible by }y \\ 0, &\text{otherwise.} \end{cases}$$
How can I define $d(x,y)$ in terms of just the basic primitive recursive functions (zero, successor, identity, projection) and the composition and primitive recursive operations?
On
Note, that the following functions are primitive recursive
Note, also, that if $f(x,\vec{u})$ is a primitive recursive function, then so is the function $y \mapsto \sum_{x\leq y} f(x,\vec{u})$.
Using this we can define the divisibility predicate by $$d ( x , y ) = \textstyle{\sum_{z \leq x}} \mathrm{Eq} ( y \cdot z , x ).$$ (In actual fact, this equation is not that cryptic. Note that if $\chi(y,\vec{x})$ is the characteristic function of some predicate, then $\sum_{y \leq z} \chi(y,\vec{x})$ will be positive iff $\chi(y,\vec{x}) = 1$ for some $y \leq z$, and so $\min \{ 1 , \sum_{y \leq z} \chi(y,\vec{x}) \}$ will be the characteristic function for the predicate "$(\exists y \leq z) ( \chi(y,\vec{x}))$." Since given $x,y$ there is at most one $z \leq x$ such that $y \cdot z = x$, we can dispense with the "$\min$" in this specific case. The right-hand-side of the formula above can then be read as the characteristic function of the predicate "$( \exists z \leq x ) ( y \cdot z = x )$" which is exactly what we mean by "$x$ is divisible by $y$.")
The natural approach would be to define the modulus function $$ m(0,y) = 0 $$ $$ m(x+1,y) = \begin{cases} 0 & \text{if }m(x,y)+1= y \\ m(x,y)+1 &\text{otherwise}\end{cases}$$ Then $d$ is simply $$ d(x,y) = \begin{cases} 1 &\text{if }m(y,x)=0 \\ 0 &\text{otherwise}\end{cases}$$
So everything will be easy if you can express $$ f(a,b,x,y) = \begin{cases} a & \text{if }x=y \\ b & \text{otherwise} \end{cases} $$ Do you have some components that might be useful for that? For example if you have the restricted subtraction function $(x,y)\mapsto \max(0,x-y)$, you can build $|x-y|$ from that, and then implement $f$ by primitive recursion over the value of $|x-y|$...