I predict that someone such as Steve Lack or Mike Shulman will tell you about the existence of (co)limits in Mon, and they'll do it better than I would, so instead I'll address a question in the last paragraph: do $M(0)$, $M(1)$ and $M(0 \to 1)$ tell you much about the rest of $M$? The answer is basically no.
To see this -- and to understand monads -- it's helpful to observe that if $M$ is regarded as an algebraic theory then $M(n)$ is the set of words in $n$ letters, or equivalently $n$-ary operations in the theory. For example, if $M$ is the monad for groups then $M(n)$ is the set of words-in-the-group-theory-sense in $n$ letters, which are the same as the $n$-ary operations in the theory of groups. For example, $x^3 y^2 x^{-1}$ is a typical word in two letters, and $(x, y) \mapsto x^3 y^2 x^{-1}$ is a typical binary operation (way of turning a pair of elements of a group into a single element). Similarly, if $M$ is the monad for rings then $M(n)$ is the set of polynomials over $\mathbb{Z}$ in $n$ variables, which are the $n$-ary operations in the theory of rings.
(Personally I'm happy to think of any monad on Set as an algebraic theory, although others prefer a more restrictive definition of theory. Anyway, this point of view doesn't matter in what follows.)
In particular, $M(0)$ is the set of nullary operations, or constants, in the theory, and $M(1)$ is the set of unary operations. Now let's consider on the one hand the identity monad $I$, and on the other the monad $M$ corresponding to the theory generated by a single binary operation $\cdot$ subject to the equation $x\cdot x = x$. (Or if you want a more familiar but more complicated example, take $M$ to be the monad for [not necessarily bounded] lattices or semilattices; the point is that $x \vee x = x$.) We then have $I(0) = 0 = M(0)$ and $I(1) = 1 = M(1)$, hence also
$$
I(0 \to 1) = (0 \to 1) = M(0 \to 1).
$$
But $I$ and $M$ are very different monads.
On the positive side, I think you might be interested in the following result; it seems in the spirit of your question. Suppose there exists an $M$-algebra with more than one element. Then:
- $M$ reflects isomorphisms
- $M$ is faithful
- the unit $\eta$ of $M$ is monic (i.e. the free $M$-algebra on a set of generators contains the generators as a subset -- it doesn't identify any of them).
What's more, all but two monads on Set have this property that there exists an algebra with more than one element. One of the exceptions is the monad $M$ with $M(A) = 1$ for all sets $A$; it's the theory generated by a single constant $e$ and the equation $x = e$. The other is the monad $M$ with $M(A) = 1$ for all nonempty sets $A$ and $M(0) = 0$; that's the theory generated by no operations and the equation $x = y$.
Edit David asked where a proof of this "positive" result could be found. I learned it from a Cambridge Part III problem sheet by Peter Johnstone, and it's probably out there somewhere in the literature (e.g. maybe in Johnstone's Sketches of an Elephant). But since consulting the literature means getting up from the couch, I'll type out a proof instead.
So, let $M = (M, \eta, \mu)$ be a monad on Set such that there exists at least one $M$-algebra with more than one element. Write $F: \mathrm{Set} \to \mathrm{Set}^M$ for the free algebra functor, and $U: \mathrm{Set}^M \to \mathrm{Set}$ for the underlying set functor; thus, $M = UF$.
First we show that $\eta: id \to M$ is monic. This means for each set $A$ and each $a, b \in A$ with $a \neq b$, we have $\eta_A(a) \neq \eta_A(b)$. By the universal property of $\eta_A$, this is equivalent to the assertion that for some $M$-algebra $X$ and map $f: A \to U(X)$ of sets, we have $f(a) \neq f(b)$. (I'd like to draw a commutative triangle to illustrate this.) Choose any $M$-algebra $X$ with more than one element: then such an $f$ can indeed be constructed.
Next we show that $F$ is faithful. It will follow that $M = UF$ is faithful, since $U$ (being monadic) is also faithful. Faithfulness of $F$ is in fact equivalent to each $\eta_A$ being monic, by general properties of adjunctions; I think that's in Categories for the Working Mathematician. Anyway, the proof is that if $f, g: A \to B$ are maps of sets with $Ff = Fg$ then $\overline{Ff} = \overline{Fg}$ (where the bar indicates transpose); but $\overline{Ff} = \eta_B \circ f: A \to UF(B)$ and similarly for $g$, and $\eta_B$ is monic, so $f = g$.
Finally we show that $F$ reflects isos. It will follows that $M = UF$ reflects isos, since $U$ (being monadic) also reflects isos. A faithful functor always reflects both epis and monos, and any map in Set that's both epi and mono is an iso. Hence any faithful functor out of Set reflects isos, and the result follows from the previous part.
Lawvere theories can be thought of as "cartesian operads." That is, we have an analogy
$$\text{Lawvere theories} : \text{cartesian monoidal categories} :: \text{operads} : \text{symmetric monoidal categories}.$$
Consider the 2-category of symmetric monoidal cocomplete categories (where the monoidal structure distributes over colimits in both variables), or SMCCs for short, which you can think of as a flavor of categorified commutative rings (where colimits categorify addition and the symmetric monoidal structure categorifies multiplication). Examples include quasicoherent sheaves on some scheme or stack, in particular modules over a commutative ring.
The free such thing on a point is the symmetric monoidal category $S = \text{Psh}(\text{FinSet}^{\times})$ of combinatorial species equipped with the symmetric monoidal structure given by Day convolution starting from disjoint union, which you can think of as a categorified polynomial / power series ring. (Exercise: show that $\text{FinSet}^{\times}$ with disjoint union is the free symmetric monoidal category on a point.)
Now, if $C$ is any other SMCC, we have an equivalence $C \cong [S, C]$ of categories (where $[-, -]$ denotes homs in this 2-category: symmetric monoidal cocontinuous functors). It follows that every SMCC admits a canonical action by the monoid $[S, S] \cong S$ under composition, which is again combinatorial species, but now equipped with a new (no longer symmetric) monoidal structure, the sustitution product, which categorifies composition of power series.
Observation: An operad is precisely a monoid in $S$ with respect to the substitution product.
This is a nice exercise. The significance of this observation is that the action of $S$ on every SMCC means that monoids in $S$ correspond to natural families of monads acting on every SMCC: explicitly, if $O_n$ is an operad, the corresponding family of monads acting on every SMCC has underlying functor the corresponding "power series"
$$X \mapsto \bigsqcup_n O_n \times_{S_n} X^{\otimes n}.$$
Hence:
Slogan: Operads are natural families of monads on SMCCs.
Now you can tell exactly the same story with "monoidal" instead of "symmetric monoidal," and you will get not-necessarily-symmetric operads. You can also tell exactly the same story with "cartesian monoidal" instead of "symmetric monoidal," and you will get Lawvere theories. In more detail:
Now consider the 2-category of cartesian monoidal cocomplete categories (where products distribute over colimits in both variables), or CMCCs for short. Examples include any Grothendieck topos. The free such thing on a point is the cartesian monoidal category $F = \text{Psh}(\text{FinSet}^{op})$, which for the same reasons as above also has a substitution product. (Exercise #1: show that $\text{FinSet}^{op}$ is the free cartesian monoidal category on a point.) (Exercise #2: show that $\text{Psh}(\text{FinSet}^{op}) \cong [\text{FinSet}, \text{Set}]$, with the substitution product, is monoidally equivalent to the monoidal category of finitary endofunctors of $\text{Set}$, with the composition product.)
Observation: A Lawvere theory is precisely a monoid in $F$ with respect to the substitution product.
As above, $F$ acts on every CMCC, so Lawvere theories correspond to natural families of monads acting on every CMCC: explicitly, if $L_n$ is a Lawvere theory, the corresponding family of monads has underlying functor
$$X \mapsto \int^n L_n \times X^n$$
where the coend is indexed over the category of finite sets. (The above formula for operads is also a coend.) Hence:
Slogan: Lawvere theories are natural families of monads on CMCCs.
You might object a little to this story because you can talk about algebras over operads and models of Lawvere theories without requiring the existence of colimits, just a symmetric monoidal or cartesian monoidal structure respectively. But this isn't much of an issue in practice: just pass to the presheaf category.
Todd Trimble has written about this circle of ideas on the nLab, for example here.
This story admits very clean generalizations. For example you can try talking about braided monoidal cocomplete categories and you will get a notion of "braided operad"; you can enrich everything in sight; generally you can talk about natural families of endofunctors on any kind of category with extra structure (these correspond to endomorphisms of the forgetful functor from categories-with-extra-structure to categories), and from there you can talk about natural families of monads.
Best Answer
As far as I know, the first paper on this was:
He called them monad functors. For the same thing but with the direction of the natural transformation reversed, he called them monad opfunctors. You can also consider the case where the natural transformation is an isomorphism, or even the identity, but I forget what he called them.
(If Street's wasn't the first paper on this, it was certainly early and very influential. And it's a beautiful paper.)
In a bid to try to make the terminology of 2-category theory more systematic, I called them lax maps of monads in my book Higher Operads, Higher Categories (CUP, 2004), and similarly colax etc. I don't know whether anyone else followed suit.
Recently I examined the thesis of one of Street's students, who I think used different terminology from what's in The formal theory of monads. I forget what it was, but I'll look it up if I remember.