The question was about intuitionism specifically, not some variant of constructivism, nor about some particular formalization of intuitionism (I don't think an intuitionist would recognize any particular formalization as being complete or even meaningful).

Your statement is not of the form $$P \to (Q \vee R).$$ It's of the form $$(\forall n)\Big(P(n) \to \big(Q(n) \vee R(n)\big)\Big).$$ (Quantification here is over natural numbers.)

To prove this intuitionistically, we don't necessarily need a proof of $(\forall n)(P(n) \to Q(n))$ or a proof of $(\forall n)(P(n) \to R(n)).$ What we need is a constructive way of finding, for each natural number $n,$ either a proof for that specific $n$ of $P(\underline{n}) \to Q(\underline{n})$ or a proof for that specific $n$ of $P(\underline{n}) \to R(\underline{n}),$ where $\underline{n}$ is the numeral representing $n.$

In general, to prove intuitionistically that $(\forall n)S(n),$ we need a constructive way of finding, for each natural number $n,$ a proof of $S(\underline{n})$ for that specific $n.$

In your example, it's clear that one can intuitionistically determine whether $b$ is composite or prime (simply check all possible factors between $2$ and $b-1\text{)}.$ If $b$ is composite, we immediately have a proof that "$b$ is composite or $(b \mid a)\text{."}$ If $b$ is prime, then since we are given that $n\ne 1\,\wedge\,(n\mid a)\,\wedge\,(n\mid b),$ we can conclude that $n=b,$ so $b\mid a,$ and again we have a proof of "$b$ is composite or $(b \mid a)\text{."}$

So we have an intuitionistically acceptable method, given any any $a, b, \text{ and }n$ such that $n\ne 1$ and $n$ divides both $a$ and $b$, of finding a proof that $"\!\underline{b}$ is a composite number or $\underline{b}$ divides $\underline{a}\!",$ which is exactly what is needed.

Now, there are ultrafinitists who might dispute the fact that each natural number $\gt 1$ is either prime or composite, but intuitionists would have no problem with that statement.

The question was about intuitionism specifically, not some variant of constructivism, nor about some particular formalization of intuitionism (I don't think an intuitionist would recognize any particular formalization as being complete or even meaningful).

Your statement is not of the form $$P \to (Q \vee R).$$ It's of the form $$(\forall n)\Big(P(n) \to \big(Q(n) \vee R(n)\big)\Big).$$ (Quantification here is over natural numbers.)

To prove this intuitionistically, we don't necessarily need a proof of $(\forall n)(P(n) \to Q(n))$ or a proof of $(\forall n)(P(n) \to R(n)).$ What we need is a constructive way of finding, for each natural number $n,$ either a proof for that specific $n$ of $P(\underline{n}) \to Q(\underline{n})$ or a proof for that specific $n$ of $P(\underline{n}) \to R(\underline{n}),$ where $\underline{n}$ is the numeral representing $n.$

In general, to prove intuitionistically that $(\forall n)S(n),$ we need a constructive way of finding, for each natural number $n,$ a proof of $S(\underline{n})$ for that specific $n.$

In your example, it's clear that one can intuitionistically determine whether $b$ is composite or prime (simply check all possible factors between $2$ and $b-1\text{)}.$ If $b$ is composite, we immediately have a proof that "$b$ is composite or $(b \mid a)\text{."}$ If $b$ is prime, then since we are given that $n\ne 1\,\wedge\,(n\mid a)\,\wedge\,(n\mid b),$ we can conclude that $n=b,$ so $b\mid a,$ and again we have a proof of $\text{"}b$ is composite or $(b \mid a)\text{."}$

So we have an intuitionistically acceptable method, given any any $a, b, \text{ and }n$ such that $n\ne 1$ and $n$ divides both $a$ and $b$, of finding a proof that $"\!\underline{b}$ is a composite number or $\underline{b}$ divides $\underline{a}\!",$ which is exactly what is needed.

Now, there are ultrafinitists who might dispute the fact that each natural number $\gt 1$ is either prime or composite, but intuitionists would have no problem with that statement.