This was supposed to be a comment on Neil's answer but it was too long.

Here is a more general class of tree that has a monoid structure. Let $X$ be a tree. Consider the tree obtained by considering only internal vertices (vertices with valence greater than 1) and the edges between them. Assume that this subtree has no vertices with valence greater than 2. Then there is a monoid structure on $X$.

Specifically, this property means that there is a sequence of vertices $v_1,\ldots, v_n$ so that every vertex is either one of the $v_i$ or has valence one and shares its unique edge with one of the $v_i$. Further, $v_i$ shares an edge with $v_{i+1}$. For $i \lt n$, let $r_i$ be one less than the valence of $v_i$. For $i=n$, let $r_n$ be the valence of $v_n$. This number is at least $1$. Let $A_i$ be Neil's monoid $\{z\in \mathbb{C}: z^{r_i}\in [0,1]\}$. Consider the monoid $A_1\times\cdots \times A_n$, and consider the subsets $M_i=\{0\}\times\cdots\times\{0\}\times A_i\times\{1\}\times \cdots \times\{1\}$. $M_1$ contains the unit of the product; $M_i$ is closed under the monoidal product, and if $i\lt j$, then $M_i\times M_j\subset M_j$.

Now let $M$ be the union of the $M_i$. topologically, $M_i$ is $A_i$, that is, a corolla with $r_i$ edges. The corolla $M_i$ is glued to $M_{i+1}$ along the point $(\underbrace{0,\ldots, 0}_i, 1,\ldots, 1)$, which is the central vertex of $M_i$ and an extremal vertex of $M_{i+1}$. This yields $X$.

Note that not every boundary point of $X$ arises as the unit of a monoid structure under this construction.

Edit:

My answer to mathoverflow.net/questions/91327 shows that the universal cover of the theta graph (and many similar trees) cannot carry a monoid structure.

This was supposed to be a comment on Neil's answer but it was too long.

Here is a more general class of tree that has a monoid structure. Let $X$ be a tree. Consider the tree obtained by considering only internal vertices (vertices with valence greater than 1) and the edges between them. Assume that this subtree has no vertices with valence greater than 2. Then there is a monoid structure on $X$.

Specifically, this property means that there is a sequence of vertices $v_1,\ldots, v_n$ so that every vertex is either one of the $v_i$ or has valence one and shares its unique edge with one of the $v_i$. Further, $v_i$ shares an edge with $v_{i+1}$. For $i \lt n$, let $r_i$ be one less than the valence of $v_i$. For $i=n$, let $r_n$ be the valence of $v_n$. This number is at least $1$. Let $A_i$ be Neil's monoid $\{z\in \mathbb{C}: z^{r_i}\in [0,1]\}$. Consider the monoid $A_1\times\cdots \times A_n$, and consider the subsets $M_i=\{0\}\times\cdots\times\{0\}\times A_i\times\{1\}\times \cdots \times\{1\}$. $M_1$ contains the unit of the product; $M_i$ is closed under the monoidal product, and if $i\lt j$, then $M_i\times M_j\subset M_j$.

Now let $M$ be the union of the $M_i$. topologically, $M_i$ is $A_i$, that is, a corolla with $r_i$ edges. The corolla $M_i$ is glued to $M_{i+1}$ along the point $(\underbrace{0,\ldots, 0}_i, 1,\ldots, 1)$, which is the central vertex of $M_i$ and an extremal vertex of $M_{i+1}$. This yields $X$.

Note that not every boundary point of $X$ arises as the unit of a monoid structure under this construction.

Edit:

My answer to mathoverflow.net/questions/91327 shows that the universal cover of the theta graph (and many similar trees) cannot carry a monoid structure.

This was supposed to be a comment on Neil's answer but it was too long.

Here is a more general class of tree that has a monoid structure. Let $X$ be a tree. Consider the tree obtained by considering only internal vertices (vertices with valence greater than 1) and the edges between them. Assume that this subtree has no vertices with valence greater than 2. Then there is a monoid structure on $X$.

Specifically, this property means that there is a sequence of vertices $v_1,\ldots, v_n$ so that every vertex is either one of the $v_i$ or has valence one and shares its unique edge with one of the $v_i$. Further, $v_i$ shares an edge with $v_{i+1}$. For $i \lt n$, let $r_i$ be one less than the valence of $v_i$. For $i=n$, let $r_n$ be the valence of $v_n$. This number is at least $1$. Let $A_i$ be Neil's monoid $\{z\in \mathbb{C}: z^{r_i}\in [0,1]\}$. Consider the monoid $A_1\times\cdots \times A_n$, and consider the subsets $M_i=\{0\}\times\cdots\times\{0\}\times A_i\times\{1\}\times \cdots \times\{1\}$. $M_1$ contains the unit of the product; $M_i$ is closed under the monoidal product, and if $i\lt j$, then $M_i\times M_j\subset M_j$.

Now let $M$ be the union of the $M_i$. topologically, $M_i$ is $A_i$, that is, a corolla with $r_i$ edges. The corolla $M_i$ is glued to $M_{i+1}$ along the point $(\underbrace{0,\ldots, 0}_i, 1,\ldots, 1)$, which is the central vertex of $M_i$ and an extremal vertex of $M_{i+1}$. This yields $X$.

Note that not every boundary point of $X$ arises as the unit of a monoid structure under this construction.

This was supposed to be a comment on Neil's answer but it was too long.

Here is a more general class of tree that has a monoid structure. Let $X$ be a tree. Consider the tree obtained by considering only internal vertices (vertices with valence greater than 1) and the edges between them. Assume that this subtree has no vertices with valence greater than 2. Then there is a monoid structure on $X$.

Specifically, this property means that there is a sequence of vertices $v_1,\ldots, v_n$ so that every vertex is either one of the $v_i$ or has valence one and shares its unique edge with one of the $v_i$. Further, $v_i$ shares an edge with $v_{i+1}$. For $i \lt n$, let $r_i$ be one less than the valence of $v_i$. For $i=n$, let $r_n$ be the valence of $v_n$. This number is at least $1$. Let $A_i$ be Neil's monoid $\{z\in \mathbb{C}: z^{r_i}\in [0,1]\}$. Consider the monoid $A_1\times\cdots \times A_n$, and consider the subsets $M_i=\{0\}\times\cdots\times\{0\}\times A_i\times\{1\}\times \cdots \times\{1\}$. $M_1$ contains the unit of the product; $M_i$ is closed under the monoidal product, and if $i\lt j$, then $M_i\times M_j\subset M_j$.

Now let $M$ be the union of the $M_i$. topologically, $M_i$ is $A_i$, that is, a corolla with $r_i$ edges. The corolla $M_i$ is glued to $M_{i+1}$ along the point $(\underbrace{0,\ldots, 0}_i, 1,\ldots, 1)$, which is the central vertex of $M_i$ and an extremal vertex of $M_{i+1}$. This yields $X$.

Note that not every boundary point of $X$ arises as the unit of a monoid structure under this construction.

Edit:

My answer to mathoverflow.net/questions/91327 shows that the universal cover of the theta graph (and many similar trees) cannot carry a monoid structure.