We shall first define a functor
$$
F:\mathsf{Mon}^{\text{op}}\to\mathsf{CRing},
$$
where $\mathsf{Mon}^{\text{op}}$ is the category opposite to the category of monoids and $\mathsf{CRing}$ is the category of commutative rings with one.
Let $\mathsf{CSRing}$ be the category of commutative semirings with one, and $L:\mathsf{CSRing}\to\mathsf{CRing}$ the left adjoint to the forgetful functor (in particular $L(\mathbb N)=\mathbb Z$).
Our functor $F$ will be the composite $L\circ F'$, where $F':\mathsf{Mon}^{\text{op}}\to\mathsf{CSRing}$ is defined as follows:
Let $M$ be a monoid, $M$–$\mathsf{Set}_{\text{fin}}$ the category of finite $M$-sets, and $$
S\subset M\text{-}\mathsf{Set}_{\text{fin}}
$$
a skeleton. Set $0:=\varnothing\in S$, let $1\in S$ be the terminal object, and for $X,Y\in S$ let $X+Y\in S$ be the coproduct of $X$ and $Y$, and $XY\in S$ the product of $X$ and $Y$.
It is straightforward to check that the formula $F'(M):=S$ defines a functor $F':\mathsf{Mon}^{\text{op}}\to\mathsf{CSRing}$, and we can set $F:=L\circ F'$.
As $F$ sends the trivial monoid to $\mathbb Z$, we have a natural morphism $F(M)\to\mathbb Z$. In other words, it would be better to view $F$ as a (contravariant) functor from monoids to commutative rings over $\mathbb Z$. In particular $F(M)$ contains $\mathbb Z$.
Question 1. Is the Krull dimension of $F(M)$ always equal to one?
Question 2. If the monoid $M$ is a group $G$, is $F(G)$ always integral over $\mathbb Z$?
The answer is Yes if
$\bullet$ the finite index subgroups of $G$ are normal,
or if
$\bullet\ G$ is finite.
(See below.)
Clearly, if $M$ is a group $G$, and $f:G\to\hat G$ is the morphism to the profinite completion, then $Ff:F(\hat G)\to F(G)$ is an isomorphism.
The ring $F(M)$ can be described as follows:
Say that an $M$-set is indecomposable if it cannot be written as a disjoint union of sub-$M$-sets in a nontrivial way, and let $I$ be the set of indecomposable objects of cardinality at least two in our skeleton $S$. For all $X,Y\in I$ we have
$$
XY=\sum_{Z\in I}\ c_{XY}^Z\ Z,
$$
where each $(c_{XY}^Z)_{Z\in I}$ is a finitely supported family of nonnegative integers. We get
$$
F(M)\simeq\mathbb Z\left[(T_X)_{X\in I}\right]/\mathfrak a,
$$
where the $T_X$ are indeterminates and $\mathfrak a$ is the ideal generated by the
$$
T_XT_Y-\sum_{Z\in I}\ c_{XY}^Z\ T_Z.
$$
In particular the element $1\in F(M)$ and the images $t_X$ of the $T_X$ form a $\mathbb Z$-basis of $F(M)$, and the natural morphism $F(M)\to\mathbb Z$ sends $t_X$ to zero.
Also note that if $M$ is a group $G$, and $N$ a normal subgroup of index $i<\infty$, then we have $(t_{G/N})^2=it_{G/N}$, and $t_{G/N}$ is integral over $\mathbb Z\subset F(G)$. This justifies the first claim after Question 2.
To prove the second claim after Question 2, recall that $I$ is the set of indecomposable objects of cardinality at least two in the skeleton $S$, and that $F(G)$ is generated by a family $(t_X)_{X\in I}$.
Assume first that the monoid $M$ is a (possibly infinite) group $G$. Order $I$ by setting $X\le Y$ if there is a surjective morphism $Y\to X$. We claim
(a) for all $X\in I$ the ring $F(G)$ is integral over the subring generated by the $t_Y$ with $Y>X$.
More precisely:
(b) for all $X\in I$ we have
$$
X^2=nX+\sum_{Y>X}n_YY
$$
with $n,n_Y\in\mathbb N$.
To prove (b) note that, for all $x_1,x_2\in X$, the stabilizer $H$ of $(x_1,x_2)\in X^2$ is the intersection of the stabilizers $H_1$ and $H_2$ of $x_1$ and $x_2$. Thus we have either $H=H_1=H_2$ and $G(x_1,x_2)\simeq X$ , or $H<H_i$ for $i=1,2$, and $G(x_1,x_2)>X$ (more correctly $Y>X$ if $Y$ is the unique element of $I$ isomorphic to $G(x_1,x_2)$). This proves (b), and thus (a).
Clearly, if $G$ is finite, (a) implies that $F(G)$ is integral over $\mathbb Z$, which is the second claim after Question 2.
Here are some examples:
As already indicated, if $M$ is the trivial monoid, then $F(M)\simeq\mathbb Z$. We also have $F(\mathbb Q)\simeq\mathbb Z$.
The ring $F(\mathbb Z)$ admit a $\mathbb Z$-basis $\{1,t_2,t_3,\dots\}$ with
$$
t_it_j=(i\land j)\ t_{i\lor j},
$$
where $i\land j$ and $i\lor j$ denote the gcd and the lcm of $i$ and $j$.
If $M$ is the monoid $\{0,1\}$ with the obvious multiplication, then the ring $F(M)$ admit a $\mathbb Z$-basis $\{1,t_1,t_2,\dots\}$ with
$$
t_it_j=t_{(i+1)(j+1)-1}.
$$
If $S_3$ denotes the symmetric group on three letters, then the ring $F(S_3)$ admit a $\mathbb Z$-basis $\{1,t_2,t_3,t_6\}$ with
$$
t_2^2=2t_2,\quad t_3^2=t_3+t_6,\quad t_it_6=it_6,\quad t_2t_3=t_6.
$$
If $G$ is the Klein four-group (that is, the non-cyclic group of order $4$), then the ring $F(G)$ admit a $\mathbb Z$-basis $\{1,t_1,t_2,t_3,u\}$ with
$$
t_i^2=2t_i,\quad u^2=4u,\quad t_iu=2u,\quad t_it_j=u\text{ for }i\ne j.
$$
Best Answer
First, a remark: your construction is known as the Burnside ring (usually considered only in the case of a group) which I imagine will aid you in finding more information about it.
Question 1: No. Indeed, the monoid $M=\{0,1\}$ you mentioned is a counterexample. Let me write $x_n$ for what you have written $t_{n-1}$, so $x_n$ is the indecomposable $M$-set with $n$ elements (the $n$-element set with only one point in the image of $0$). Then these elements $x_n$ satisfy $x_nx_m=x_{nm}$. This makes it clear that actually $F(M)$ is just the polynomial ring $\mathbb{Z}[x_2,x_3,x_5,\dots]$ on the elements $x_p$ where $p$ is prime. This ring has infinite Krull dimension.
Question 2: Yes. To prove this, let $X$ be any finite $G$-set and let $K$ be the kernel of the action of $G$ on $X$. Note that $K$ is a finite index normal subgroup of $G$, and acts trivially on $X^n$ for all $n$. It follows that actually the subring of $F(G)$ generated by $X$ is isomorphic to the subring of $F(G/K)$ generated by $X$. Since $G/K$ is finite, this shows $X$ is integral over $\mathbb{Z}$ by the work you have done.
Incidentally, there is a quicker way to see $F(G)$ is integral over $\mathbb{Z}$ when $G$ is a finite group. Just note that $F(G)$ is generated as an abelian group by the $G$-sets $G/H$ where $H$ ranges over all subgroups of $G$. In particular, $F(G)$ is a finitely generated $\mathbb{Z}$-module and thus is integral over $\mathbb{Z}$.