Formal power series: are subgroups of the multiplicative group closed? I have a subgroup $G$ of the multiplicative subgroup of the ring of formal power series on n coefficients, $(\mathbb{Z}[[X_1,...,X_n]])^*$, and an element $a = \sum_{i \in I} a_i X_1^{i_1} ... X_n^{i_n}$ ($I$ a set of multi-indices).
Now assume that all finite-degree sums in $a$ belong to $G$ (for any $n$, the sum of the summands of $a$ of degree lesser than $n$ is in $G$). Is $a$ in $G$? And if this isn't always the case, a counter-example would be nice.
A slightly more general question, which is where the title comes from, is whether/which multiplicative subgroups are closed in the usual topology on $\mathbb{Z}[[X_1,...,X_n]]$.

Following is a counterexample to your first question in $\mathbb{Z}[[x]]$. 
The main idea is that second order linearly recurrent sequences will bear witness (NB: geometric sequences will not!). Hopefully this will help you attack the second question, too. 

Let $a = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + 6x^5 + \cdots \in \mathbb{Z}[[x]]$.  Let $G$ be the subgroup of $\mathbb{Z}[[x]]^*$ generated by the 'finite sums' of $a$ as you have defined them in your post.  Then $a \notin G$.  

Proof:
Any element of $G$ is a finite product of the elements $1, 1+2x, (1+2x)^{-1}, 1+2x + 3x^2, (1+2x + 3x^2)^{-1}, \ldots$.  
It follows that for any $b \in G$, there exists a polynomial $$f=(1+2x)^{n_1}(1+2x + 3x^2)^{n_2}(1+2x + 3x^2 + 4x^3)^{n_3}\cdots$$ (where finitely many $n_i$ are not equal to $1$) such that $fG \in \mathbb{Z}[x]$.  
Define $I_b = \{g \in \mathbb{Q}[x] \mid gb \in \mathbb{Q}[x]\}$, which is an ideal of $\mathbb{Q}[x]$.  Since $\mathbb{Q}[x]$ is a principal ideal domain,  it is easy to see that this ideal is principally generated by $a^{-1} = 1 - 2x + x^2 = (1 - x)^2$.  
Therefore if $b \in G$ we see that $(1 - x)^2$ must divide, in $\mathbb{Q}[x]$, a polynomial of the form $f$ given above.
But this is plainly absurd because it would imply that $f$ has $x = 1$ as a root, whereas $f(1) > 0$ is apparent.  $\square$
