This is a partial answer to YCor's improved question about an identity between associativity and constancy of triples, reducing it to a finite check. (I'll use multiplicative notation below, for simplicity.)

Consider the multiplication on the set $\{a,b,c,d,e,0\}$ where $ab=d$, $dc=e$, and all other products are $0$. This algebra satisfies any identity where both sides involve four or more products (since such products are $0$). Moreover, this algebra is not associative, since $a(bc)\neq (ab)c$. So the identity we are looking for must have at least one side that involves no more than 3 products.

Now consider the multiplication on the set $\{a,b,0\}$ where $aa=b$, and all other products are $0$. This algebra satisfies constancy of triples. Any identity where one side has three or more products and the other side has fewer than three products is not satisfied in this algebra. So the identity we are looking for must have at least one side involving exactly 3 products, and the other side must involve at least 3 products.

Write such an identity as $t=t'$, where $t$ involves 3 products exactly. Thus, up to symmetry and renaming, $t$ is of the form $x(yz)$, or $x(yy)$, or $x(yx)$, or $x(xy)$, or $x(xx)$.

It is easy to check that all five options are always $0$ in the first algebra we constructed above. So, if $t'$ involves four or more products, then $t=t'$ is satisfied in the first algebra we constructed, but associativity doesn't follow.

Thus, $t'$ also involves exactly 3 products. In order to avoid $t=t'$ being again satisfied in the first algebra, we must have $t'$ of the form $(pq)r$, where $p,q,r$ are distinct variables. These variables needn't be distinct from $x,y,z$.

We are now reduced to a finite check.

This is a partial answer to YCor's improved question about an identity between associativity and constancy of triples, reducing it to a finite check. (I'll use multiplicative notation below, for simplicity.)

Consider the multiplication on the set $\{a,b,c,d,e,0\}$ where $ab=d$, $dc=e$, and all other products are $0$. This algebra satisfies any identity where both sides involve four or more letters (equivalently, 3 or more products) since such products are $0$. Moreover, this algebra is not associative, since $a(bc)\neq (ab)c$. So the identity we are looking for must have at least one side that involves no more than 3 letters.

Now consider the multiplication on the set $\{a,b,0\}$ where $aa=b$, and all other products are $0$. This algebra satisfies constancy of triples. Any identity where one side has three or more letters and the other side has fewer than three letters is not satisfied in this algebra. So the identity we are looking for must have at least one side involving exactly 3 letters and the other side must involve at least 3 letters OR both sides involve two letters or less.

I believe the second option can be handled very easily, but in any case it is a finite check. So, consider the first option. Write such an identity as $t=t'$, where $t$ involves 3 letters exactly. Thus, up to symmetry and renaming, $t$ is of the form $x(yz)$, or $x(yy)$, or $x(yx)$, or $x(xy)$, or $x(xx)$.

It is easy to check that all five options are always $0$ in the first algebra we constructed above. So, if $t'$ involves four or more letters, then $t=t'$ is satisfied in the first algebra we constructed, but associativity doesn't follow.

Thus, $t'$ also involves exactly 3 letters. In order to avoid $t=t'$ being again satisfied in the first algebra, we must have $t'$ of the form $(pq)r$, where $p,q,r$ are distinct variables. These variables needn't be distinct from $x,y,z$.

We are now reduced to a finite check.

This is a long comment, that reduces YCor's improved question about an identity between associativity and constancy of triples to a finite check. (I'll use multiplicative notation below, for simplicity.)

Consider the multiplication on the set $\{a,b,c,d,e,0\}$ where $ab=d$, $dc=e$, and all other products are $0$. This algebra satisfies any identity where both sides involve four or more products (since such products are $0$). Moreover, this algebra is not associative, since $a(bc)\neq (ab)c$. So the identity we are looking for must have at least one side that involves no more than 3 products.

Now consider the multiplication on the set $\{a,b,0\}$ where $aa=b$, and all other products are $0$. This algebra satisfies constancy of triples. Any identity where one side has three or more products and the other side has fewer than three products is not satisfied in this algebra. So the identity we are looking for must have at least one side involving exactly 3 products, and the other side must involve at least 3 products.

Write such an identity as $t=t'$, where $t$ involves 3 products exactly. Thus, up to symmetry and renaming, $t$ is of the form $x(yz)$, or $x(yy)$, or $x(yx)$, or $x(xy)$, or $x(xx)$.

It is easy to check that all five options are always $0$ in the first algebra we constructed above. So, if $t'$ involves four or more products, then $t=t'$ is satisfied in the first algebra we constructed, but associativity doesn't follow.

Thus, $t'$ also involves exactly 3 products. This means that there are now only finitely many options to consider. I would guess than none of them works out to be strictly between constancy of triples and associativity.

This is a partial answer to YCor's improved question about an identity between associativity and constancy of triples, reducing it to a finite check. (I'll use multiplicative notation below, for simplicity.)

Consider the multiplication on the set $\{a,b,c,d,e,0\}$ where $ab=d$, $dc=e$, and all other products are $0$. This algebra satisfies any identity where both sides involve four or more products (since such products are $0$). Moreover, this algebra is not associative, since $a(bc)\neq (ab)c$. So the identity we are looking for must have at least one side that involves no more than 3 products.

Now consider the multiplication on the set $\{a,b,0\}$ where $aa=b$, and all other products are $0$. This algebra satisfies constancy of triples. Any identity where one side has three or more products and the other side has fewer than three products is not satisfied in this algebra. So the identity we are looking for must have at least one side involving exactly 3 products, and the other side must involve at least 3 products.

Write such an identity as $t=t'$, where $t$ involves 3 products exactly. Thus, up to symmetry and renaming, $t$ is of the form $x(yz)$, or $x(yy)$, or $x(yx)$, or $x(xy)$, or $x(xx)$.

It is easy to check that all five options are always $0$ in the first algebra we constructed above. So, if $t'$ involves four or more products, then $t=t'$ is satisfied in the first algebra we constructed, but associativity doesn't follow.

Thus, $t'$ also involves exactly 3 products. In order to avoid $t=t'$ being again satisfied in the first algebra, we must have $t'$ of the form $(pq)r$, where $p,q,r$ are distinct variables. These variables needn't be distinct from $x,y,z$.

We are now reduced to a finite check.