I do not have reference for your type of measure, but I have a version of the Riesz representation theorem for all such measures when the lattice is distributive. In this post, we shall not assume that the mapping $\mu$ maps the least element of the lattice to the identity of the monoid.

From each lattice $(X,\wedge,\vee)$, we construct a commutative monoid $\mathbf{M}(X)$ with $X\subseteq\mathbf{M}(X)$ such that each modular map $\mu:X\rightarrow M$ extends to a monoid homomorphism $\overline{\mu}:\mathbf{M}(X)\rightarrow M$ in the usual way. When $X$ is distributive, $\mathbf{M}(X)$ is the structure monoid for a set theoretic solution to the Yang-Baxter equation.

Suppose that $T$ is a set and $T:X^2\rightarrow X^2$. Let $X^\ast$ denote the collection of all possibly empty sequences $(x_1,\dots,x_n)$ of elements in $X$. Let $\simeq$ be the smallest congruence on the monoid $X^\ast$ where
$(x,y)\simeq T(x,y)$ whenever $x,y\in X$, and let $\mathbf{M}(X,T)=X^\ast/\simeq$. We shall write $\mathbf{M}(X,\ast_1,\ast_2)$ whenever $\ast_1,\ast_2$ are the binary operations where $T(x,y)=(x\ast_1y,x\ast_2y)$, and we shall write $\simeq_T$ to specify the function $T$.

Suppose that $T(x,y)=(x\vee y,x\wedge y)$. We observe that $[x,y]=[x\vee y,x\wedge y]=[y,x]$, so the monoid $\mathbf{M}(X,T)=\mathbf{M}(X,\vee,\wedge)$ is commutative. If $\mu:X\rightarrow M$, then we extend $\mu$ to a monoid homomorphism $\overline{\mu}:\mathbf{M}(X,T)\rightarrow M$ by setting $\overline{\mu}([x_1,\dots,x_r])=\mu(x_1)+\dots+\mu(x_r)$.

Suppose that $X$ is a set and $T:X^2\rightarrow X^2$ is a function. Then we say that $T$ is a set theoretic solution to the Yang-Baxter equation if
$$(1_X\times T)\circ(T\times 1_X)\circ(1_X\times T)=(T\times 1_X)\circ(1_X\times T)\circ(T\times 1_X).$$

The motivation behind the Yang-Baxter equation is that a function $T:X^2\rightarrow X^2$ is a set theoretic solution to the Yang-Baxter equation precisely when for all $n$, we have an action of the positive braid monoid $B_n^+$ on $X^n$ defined by setting $$(x_1,\dots,x_n)\bullet\sigma_i=(x_1,\dots,x_{i-1},T(x_i,x_{i+1}),x_{i+2}\dots,x_n).$$

Whenever $b,c\in B_n^+$, there are $r,s\in B_n^+$ where $bc=rs$. As a corollary,
if $T$ is a set theoretic solution to the Yang-Baxter equation, then
$(x_1,\dots,x_n)\simeq_T(y_1,\dots,y_n)$ precisely when there are positive braids $b,c\in B_n^+$ where $(x_1,\dots,x_n)\bullet b=(y_1,\dots,y_n)\bullet c$.

Lemma: (median identity) A lattice is distributive if and only if it satisfies the following identity known as the median identity: $$(x\wedge y)\vee(x\wedge z)\vee(y\wedge z)=(x\vee y)\wedge(x\vee z)\wedge(y\vee z).$$

Proposition: In a lattice $(X,\wedge,\vee)$, the operation $(x,y)\mapsto(x\vee y,x\wedge y)$ satisfies the Yang-Baxter equation if and only if the lattice is distributive.

Proof: We have $(a,b,c)\bullet\sigma_1\sigma_2\sigma_1=(a\vee b,a\wedge b,c)\bullet\sigma_2\sigma_1=(a\vee b,(a\wedge b)\vee c,a\wedge b\wedge c)\bullet\sigma_1=(a\vee b\vee c,(a\vee b)\wedge((a\wedge b)\vee c),a\wedge b\wedge c)$.

$(a,b,c)\bullet\sigma_2\sigma_1\sigma_2=(a,b\vee c,b\wedge c)\bullet\sigma_1\sigma_2
=(a\vee b\vee c,a\wedge(b\vee c),b\wedge c)\bullet\sigma_2=(a\vee b\vee c,(b\wedge c)\vee((a\wedge(b\vee c)),a\wedge b\wedge c).$

Therefore, $(X,\wedge,\vee)$ satisfies the Yang-Baxter equation if and only if the lattice $X$ satisfies the identity $(a\vee b)\wedge((a\wedge b)\vee c)=(b\wedge c)\vee((a\wedge(b\vee c))$ which we shall call the YB-distributivity identity.

In a distributive lattice, using the median identity, we have
$$(a\vee b)\wedge((a\wedge b)\vee c)
=(a\vee b)\wedge(a\vee c)\wedge(b\vee c)
=(b\wedge c)\vee(a\wedge b)\vee(a\wedge c)
=(b\wedge c)\vee((a\wedge(b\vee c)),$$
so every distributive lattice satisfies the YB-distributivity identity.

Now, suppose that we are in a lattice that satisfies the YB-distributivity identity. Then $(a\vee b)\wedge((a\wedge b)\vee c)=(b\wedge c)\vee(a\wedge(b\vee c))=(a\vee c)\wedge((a\wedge c)\vee b)$. More generally,
$$(a_{\sigma(1)}\vee a_{\sigma(2)})\wedge((a_{\sigma(1)}\wedge a_{\sigma(2)})\vee a_{\sigma(3)})=(a_{\tau(1)}\wedge a_{\tau(2)})\vee((a_{\tau(1)}\vee a_{\tau(2)})\wedge a_{\tau(3)})$$
whenever $\sigma,\tau\in S_3$.
Thus,
$$(a_1\vee a_2)\wedge(a_1\vee a_3)\wedge(a_2\vee a_3)=\bigwedge_{\sigma\in S_3}((a_{\sigma(1)}\vee a_{\sigma(2)})\wedge((a_{\sigma(1)}\wedge a_{\sigma(2)})\vee a_{\sigma(3)}))$$
$$=\bigvee_{\tau\in S_3}((a_{\tau(1)}\wedge a_{\tau(2)})\vee((a_{\tau(1)}\vee a_{\tau(2)})\wedge a_{\tau(3)}))=(a_1\wedge a_2)\vee(a_1\wedge a_3)\vee(a_2\wedge a_3).$$

Since the lattice satisfies the median identity, the lattice must also be distributive.

Q.E.D.

In the case when $X$ is a linearly ordered set, the operation $(x,y)\mapsto(x\vee y,x\wedge y)$ is just a sorting operation, and the action of the positive braid monoid on $X$ is simply the sorting operation. Let $\delta_n=\sigma_n\dots\sigma_1$ and let $\Delta_n=\delta_1\dots\delta_{n-1}$. Then the list $(x_1,\dots,x_n)\bullet\Delta_n$ is the list obtained from $(x_1,\dots,x_n)$ by sorting $(x_1,\dots,x_n)$ from the greatest element to the least element. In particular, if $b$ is a positive braid, then
$(x_1,\dots,x_n)\bullet\Delta_n\cdot b=(x_1,\dots,x_n)\bullet\Delta_n$. Since the variety of distributive lattices is generated by the 2 element linearly ordered distributive lattice, we conclude that
$(x_1,\dots,x_n)\bullet\Delta_n\cdot b=(x_1,\dots,x_n)\bullet\Delta_n$ in the variety of all distributive lattices. As a consequence, for distributive lattices, we conclude that $(x_1,\dots,x_n)\simeq(y_1,\dots,y_n)$ if and only if $(x_1,\dots,x_n)\bullet\Delta_n=(y_1,\dots,y_n)\bullet\Delta_n$. Furthermore, in distributive lattices, for each $(x_1,\dots,x_n)\in X^*$, the element $(x_1,\dots,x_n)\bullet\Delta_n$ is the unique member $(y_1,\dots,y_n)$ of the equivalence class $[x_1,\dots,x_n]$ with $y_1\geq\dots\geq y_n$.

If $X$ is a distributive lattice, then let $\mathbf{M}^\sharp(X)$ be the monoid consisting of all (possibly empty) sequences $(x_1,\dots,x_n)$ where $x_1\geq\dots\geq x_n$ and with an operation $\bullet$ defined by letting
$$(x_1,\dots,x_m)\bullet(y_1,\dots,y_n)=(x_1,\dots,x_m,y_1,\dots,y_n)\bullet\Delta_{m+n-1}.$$
Then the monoid $\mathbf{M}^\sharp(X)$ is canonically isomorphic to the monoid $\mathbf{M}(X,\vee,\wedge)$.

Suppose now that $X$ is a set and $\mathcal{A}$ is a collection of subsets of $X$ closed under taking the union of two sets and the intersection of two sets.

Let $\mathbf{M}^!(X,\mathcal{A})$ be the collection of pairs $(f,n)$ where $n\geq 0$ and $f:X\rightarrow\\{0,\dots,n\\}$ is a function such that
$f^{-1}[\\{i,\dots,n\\}]\in\mathcal{A}$ for $1\leq i\leq n$.

Give $\mathbf{M}^!(X,\mathcal{A})$ the operation $+$ defined by letting
$(f,m)+(g,n)=(f+g,m+n)$. The operation $+$ is clearly associative and commutative.

Define a $\Gamma:\mathbf{M}^!(X,\mathcal{A})\rightarrow \mathbf{M}(\mathcal{A})$ by setting $$\Gamma((f,n))=([f^{-1}[\\{1,\dots,n\\}],\dots,f^{-1}[\\{n\\}]],n).$$

Define a function $\Delta:\mathbf{M}(\mathcal{A})\rightarrow\mathbf{M}^!(X,\mathcal{A})$ by setting
$$\Delta([A_1,\dots,A_n])=(\chi_{A_1}+\dots+\chi_{A_n},n).$$ We observe that this definition does not depend on the equivalence class that we choose.

Proposition: The functions $\Gamma,\Delta$ are inverse monoid homomorphisms.

Proof: Observe that $\chi_{f^{-1}[\\{1,\dots,n\\}]}+\dots+\chi_{f^{-1}[\\{n\\}]}=f.$

Therefore, $$\Delta\Gamma(f,n)=\Delta([f^{-1}[\{1,\dots,n\}],\dots,f^{-1}[\{n\}]],n)
=(\chi_{f^{-1}[\{1,\dots,n\}]}+\dots+\chi_{f^{-1}[\{n\}]},n)=(f,n).$$

Suppose that $\mathbf{x}\in\mathbf{M}(\mathcal{A})$. Then there are $A_1\supseteq A_2\supseteq\dots\supseteq A_n$ with $\mathbf{x}=[A_1,\dots,A_n]$. In this case,
$(\chi_{A_1}+\dots\chi_{A_n})^{-1}[\{i,\dots,n\}]=A_i$, so
$$\Gamma\Delta(\mathbf{x})=\Gamma\Delta([A_1,\dots,A_n])=\Gamma(\chi_{A_1}+\dots+\chi_{A_n},n)=[(\chi_{A_1}+\dots+\chi_{A_n})^{-1}[\\{1,\dots,n\\}],\dots,(\chi_{A_1}+\dots+\chi_{A_n})^{-1}[\\{n\\}]]=[A_1,\dots,A_n].$$

Now observe that
$$\Delta([A_1,\dots,A_m]+[B_1,\dots,B_n])=\Delta([A_1,\dots,A_m,B_1,\dots,B_n])$$

$$=(\chi_{A_1}+\dots+\chi_{A_m}+\chi_{B_1}+\dots+\chi_{B_n},m+n)
=(\chi_{A_1}+\dots+\chi_{A_m},m)+(\chi_{B_1}+\dots+\chi_{B_n},n).$$

$$=\Delta([A_1,\dots,A_m])+\Delta([B_1,\dots,B_n]).$$

Q.E.D.

If we define integration by setting $$\int_n(\chi_{A_1}+\dots+\chi_{A_n})d\mu=\mu(A_1)+\dots+\mu(A_n),$$ then
$$\overline{\mu}(\Gamma(f,n))=\int_n f d\mu$$ for all functions $f$.

**Topological monoids and lattices**

Our construction respects the topological structure on distributive lattices and monoids.

Suppose that $X$ is a topological distributive lattice and $M$ is a topological commutative monoid. This simply means that the operations $\vee,\wedge$ on $X$ and the addition operation on $M$ are continuous. We can give $X^n$ the product topology, and we can give $\bigcup_{n=0}^\infty X^n$ the disjoint union topology where $U\subseteq\bigcup_{n=0}^\infty X^n$ is open precisely when $U\cap X^n$ is open for each $n\geq 0$. Now associate $\mathbf{M}(X)$ with $\mathbf{M}^\sharp(X)$ and give $\mathbf{M}^\sharp(X)$ the topology induced by the topology on $\bigcup_{n=0}^\infty X^n$. Then one can easily verify that the mapping $\mu:X\rightarrow M$ is continuous if and only if the extension $\overline{\mu}:\mathbf{M}^\sharp(X)\rightarrow M$ is a continuous monoid homomorphism.

**Median algebras**

The least upper bound and greatest lower bounds are in symmetric positions in the modularity identity; for distributive lattices, we can even formulate the modularity identity in terms of the median operation where there is no notion of which way is up.

A median algebra is an algebraic structure $(X,m)$ that satisfies the identities $m(x,x,y)=x,m(x,y,z)=m(y,z,x),m(x,y,z)=m(y,x,z)$ and the distributivity identity
$x\wedge_a(y\wedge_bz)=(x\wedge_ay)\wedge_b(x\wedge_az)$ where we define $m(x,y,z)=x\wedge_yz$.

In a distributive lattice $(X,\wedge,\vee)$ define the ternary median operation $m$ by setting $m(x,y,z)=(x\wedge y)\vee(x\wedge z)\vee(y\wedge z)$; in this case $(X,m)$ is a median algebra.

If $(X,m)$ is a median algebra, $G$ is a semigroup, and $\mu:X\rightarrow G$ is a function, then we say that $\mu$ is modular with respect to the median operation $m$ when $\mu(m(a,x,y))\bullet\mu(m(b,x,y))=\mu(x)\bullet\mu(y)$ whenever $x=m(a,b,x),y=m(a,b,y)$. If $\mu$ is modular with respect to $m$, then the subsemigroup $\langle\mu[G]\rangle$ is commutative, so from now on, we shall assume that all semigroups are commutative, and we shall use the $+$ symbol for the semigroup operation.

Lemma: In a median algebra, if $x=m(a,b,x),y=m(a,b,y)$, then
$m(a,x,y)\vee m(b,x,y)=x\vee y$ and $m(a,x,y)\wedge m(b,x,y)=x\wedge y$.

Proof: We have
$$m(a,x,y)\vee m(b,x,y)=(a\wedge x)\vee(a\wedge y)\vee(b\wedge x)\vee(b\wedge y)\vee(a\wedge b)$$

$$=((a\vee b)\wedge(x\vee y))\vee(a\wedge b)$$

$$=[(a\vee b)\wedge x]\vee[(a\vee b)\wedge y]\vee(a\wedge b)$$

$$=(a\wedge x)\vee(b\wedge x)\vee(a\wedge y)\vee(b\wedge y)\vee(a\wedge b)$$
$$=x\vee y.$$

The identity, $m(a,x,y)\wedge m(b,x,y)=x\wedge y$, can be obtained by switching the places of the operation $\wedge,\vee$. Q.E.D

Proposition: Let $(X,\wedge,\vee)$ be a distributive lattice with median operation $m$, and let $G$ be a commutative semigroup. Let $\mu:X\rightarrow G$ be a function. Then $\mu$ is modular with respect to the median operation $m$ if and only if $\mu$ is modular with respect to the lattice operations $\wedge,\vee$.

Proof: $\rightarrow.$ Let $a=x\wedge y,b=x\vee y$. Then
$m(a,b,x)=(a\wedge x)\vee(b\wedge x)\vee(a\wedge b)=(x\wedge y)\vee x\vee(x\wedge y)=x$ and $m(a,b,y)=y$ as well. Therefore,
$\mu(x)+\mu(y)=\mu(m(a,x,y))+\mu(m(b,x,y))=\mu(x\wedge y)+\mu(x\vee y)$.

$\leftarrow$. Suppose that $x=m(a,x,y),y=m(b,x,y)$. Then
$$\mu(m(a,x,y))+\mu(m(b,x,y))=\mu(m(a,x,y)\vee m(b,x,y))+\mu(m(a,x,y)\wedge m(b,x,y))=\mu(x\vee y)+\mu(x\wedge y)=\mu(x)+\mu(y).$$

Q.E.D.

While the construction of $\mathbf{M}(X)$ is valid for median algebras, for a general median algebra $X$, I have not been able to put the elements of $\mathbf{M}(X)$ in a normal form.