Your ring $L$ is a localization of the power series rings $R= k[[x_1,\cdots,x_n]]$ at the multiplicative set $M$ of monomials in $R$.
So the prime ideals of $L$ correspond to prime ideals in $R$ which do not meet $M$. Clearly, the maximal primes are the set of biggest primes which do not contain any variable. Since $R$ is local and the maximal ideal contains all the variables, these ideals are of dimension one (this is the key difference to the Laurent polynomials case, since in that case these ideals would be maximal in the polynomial ring, so Nullstellensatz applies)
For example, when $n=2$, you get all ideals of height one which does not contains $x_1$ or $x_2$. So they are all principal ideal $(f)$ with $f$ irreducible and $f\neq x_1$, $f\neq x_2$.
It shows that the answer to your refined question is NO. The ideals $(x-y)$ and $(x-y^2)$ can not be in the same orbit.
In general one can only describe the set of maximal ideals in $L$ as follows: they are prime ideals of height $n-1$ in $S$, minus the set $(x_i, P)$, where $P$ is a prime of height $n-2$ in $k[[x_1,...x_{i-1}, x_{i+1},...,x_n]]$.
I doubt one can say more in general, since there are many open questions about affine curves.
Suppose I have a set of disjoint, nonempty sets, and I want to choose one element from each. Consider the free polynomial ring generated by all the elements of all the sets, then take the quotient by the ideal generated by $xy$ for each pair $x$ and $y$ different elements in the same set. Any prime ideal must contain all but one element from each set.
We need to show that a minimal prime ideal does not contain all the elements from any set. Then a minimal prime ideal will give us a choice function. We can take the minimal prime to be generated by the elements, since every prime ideal contains a prime ideal generated by elements. Then remove one element from a set entirely contained in the prime ideal. The ideal will still be prime, and smaller, this is a contradiction.
So minimal primes give a choice function.
Best Answer
A typical ring of dimension $2$, like $k[x,y]$, does not have that property. Indeed, if $I=(xy)$, $J=(x^2-y^2)$, then $M_I=I$, $M_J=J$, and $M_I+M_J=I+J$ contains $x^3$ and $y^3$ but not $x$ and $y$, so is not radical, so cannot be the intersection of any set of maximal ideals.
On the other hand, in a Dedekind domain, ideals of this form are exactly products of finitely many maximal ideals or the zero ideal. Scheme-theoretically, they correspond to finite sets of points on a curve or else the whole curve. This set is closed under addition, or, scheme-theoretically, intersection, so you're OK there.
Obviously, this also works in every local ring, where there is a unique maximal ideal.