There are no terms that would define a group structure (let alone a ring) in all Heyting algebras.

The first observation is that if $f(x)$ is a term-definable involution, then $f(x)=x$: by inspection of the Rieger–Nishimura lattice, we see that all other unary term-definable $f$ are either constant in Boolean algebras, or satisfy $f(x)=f(\neg\neg x)$ ($x$ is $p$ in the picture):

The Rieger–Nishimura lattice http://upload.wikimedia.org/wikipedia/commons/thumb/5/5c/Rieger-Nishimura.svg/500px-Rieger-Nishimura.svg.png

Then, assume that $+,-,0$ are terms defining a group in any Heyting algebra (where the $0$ is not necessarily the bottom of the Heyting algebra, which I will denote as $\bot$, and the group is not assumed commutative, despite the additive notation). Since $f(x)=-x$ is an involution, we must have $-x=x$, i.e., the group is of exponent $2$. Thus, $g(x)=\bot+x$ and $h(x)=\top+x$ are also involutions, hence $\bot+x=x=\top+x$, which implies $\top=\bot$, a contradiction.

There are no terms that would define a group structure (let alone a ring) in all Heyting algebras.

The first observation is that if $f(x)$ is a term-definable involution, then $f(x)=x$: by inspection of the Rieger–Nishimura lattice, we see that all other unary term-definable $f$ are either constant in Boolean algebras, or satisfy $f(x)=f(\neg\neg x)$ ($x$ is $p$ in the picture):

Then, assume that $+,-,0$ are terms defining a group in any Heyting algebra (where the $0$ is not necessarily the bottom of the Heyting algebra, which I will denote as $\bot$, and the group is not assumed commutative, despite the additive notation). Since $f(x)=-x$ is an involution, we must have $-x=x$, i.e., the group is of exponent $2$. Thus, $g(x)=\bot+x$ and $h(x)=\top+x$ are also involutions, hence $\bot+x=x=\top+x$, which implies $\top=\bot$, a contradiction.

There are no terms that would define a group structure (let alone a ring) in all Heyting algebras.

The first observation is that if $f(x)$ is a term-definable involution, then $f(x)=x$: by inspection of the Rieger–Nishimura lattice, we see that all other unary term-definable $f$ are either constant in Boolean algebras, or satisfy $f(x)=f(\neg\neg x)$.

Then, assume that $+,-,0$ are terms defining a group in any Heyting algebra (where the $0$ is not necessarily the bottom of the Heyting algebra, which I will denote as $\bot$, and the group is not assumed commutative, despite the additive notation). Since $f(x)=-x$ is an involution, we must have $-x=x$, i.e., the group is of exponent $2$. Thus, $g(x)=\bot+x$ and $h(x)=\top+x$ are also involutions, hence $\bot+x=x=\top+x$, which implies $\top=\bot$, a contradiction.

There are no terms that would define a group structure (let alone a ring) in all Heyting algebras.

The first observation is that if $f(x)$ is a term-definable involution, then $f(x)=x$: by inspection of the Rieger–Nishimura lattice, we see that all other unary term-definable $f$ are either constant in Boolean algebras, or satisfy $f(x)=f(\neg\neg x)$ ($x$ is $p$ in the picture):

The Rieger–Nishimura lattice http://upload.wikimedia.org/wikipedia/commons/thumb/5/5c/Rieger-Nishimura.svg/500px-Rieger-Nishimura.svg.png

Then, assume that $+,-,0$ are terms defining a group in any Heyting algebra (where the $0$ is not necessarily the bottom of the Heyting algebra, which I will denote as $\bot$, and the group is not assumed commutative, despite the additive notation). Since $f(x)=-x$ is an involution, we must have $-x=x$, i.e., the group is of exponent $2$. Thus, $g(x)=\bot+x$ and $h(x)=\top+x$ are also involutions, hence $\bot+x=x=\top+x$, which implies $\top=\bot$, a contradiction.