I think that (assuming say that $M$ is finite type over a field $k$) the map $M^{\times} \to M$ should be an open immersion.
Here is an attempt at a proof, hopefully not too bogus:
First of all, let's begin by carefully defining $M^{\times}$.
We have have the multiplication map $\mu:M \times M \to M,$ the identity element $e: \mathrm{Spec} k \to M$, and the tranposition $\tau: M\times M \to M \times M$ (switching factors). (Here I am assuming I have a monoid in the category of $k$-schemes, for some field $k$.)
We may form the product $\nu:=\mu\times (\mu\circ \tau): M\times M \to M\times M,$ which in terms of $S$-valued points (for any $k$-scheme $S$) maps the pair $(x,y)$ to the pair $(xy, yx).$ If we form the fibre product of $\nu$ and $e\times e: \mathrm{Spec} k \to M\times M$, we obtain a closed subscheme $X \subset M\times M$, whose $S$-points consist of pairs $(x,y)$ such that $x y = yx = e.$
Now in fact we can make $X$ a group scheme, by defining (on $S$-valued points) the multiplication $(x,y)\cdot (u,v) = (xu,vy),$ the identity element $(e,e)$, and the inverse operation $(x,y)^{-1} = (y,x)$. Intuitively, we have that $y = x^{-1}$, and this is just the product on invertible elements.
Thus we should write $X = M^{\times}$; this is the formation of the group of units in a categorical manner.
Now projection onto the first factor gives a map $X \to M$, which is a monomorphism (look at $S$-valued points, and observe by the usual proof for monoids in sets that $x$ determines $y$).
Everything so far is purely categorical.
Now we have to do some algebraic geometry. I'll assume that $M$ is finite type over $k$, and argue with Zariski tangent spaces.
Consider the Zariski tangent space to $M$ at $e$. This is equal to morphisms $\mathrm{Spec} k[\epsilon]/(\epsilon^2) \to M$ lying above the map $e: \mathrm{Spec}k \to M$, and so is naturally a monoid in the category of $k$-vector spaces, with zero acting as the identity. A standard argument then shows that the monoid structure on the Zariski tangent space structure coincides with the underlying addition (coming from the a priori vector space structure).
Now this vector space is a group under addition, and so we can lift this map $\mathrm{Spec} k[\epsilon]/(\epsilon^2)$ from a map to $M$ to a map to $M^{\times}$.
In conclusion: the map $M^{\times} \to M$ induces an isomorphism of Zariski tangent spaces the identity. Since $M^{\times}$ is homogeneous, it must do so at every (say $\overline{k}$-valued) point of its domain. (Alternatively, we could just repeat the above argument as these points.)
So $M^{\times} \to M$ is a monomorphism that induces an isomorphism on Zariski tangent spaces at every $\overline{k}$-valued point of its domain.
This should imply that $M^{\times} \to M$ is an etale monomorphism, and hence is an open immersion. (Hopefully this last step is correct. If $M$ is smooth it is okay; if $M$ is not smooth then it also seems okay to me at the moment, but perhaps I am blundering ... .)