When is $1+x+x^2+x^3+... $ and $ {1 \over 1-x} \quad (x \ne 1) $ *not* interchangeable in an algebraic formula? [update 3]: This question was stated from a wrong premise ("...fails obviously...") which I became aware of after comments and answers, and thus I tend to retract it. But there are those constructive answers, sheding light on this, so I think it is better to keep the question together with the answers alive

This is merely an accidental question, to improve my understanding of the concept of divergent summation.        
I'm nearly completely used to the assumtion, that $ \sum_{k=0}^\infty x^k = {1 \over 1-x}$ by analytic continuation can be inserted in any formula (except of course for $x \ne 1$) - maybe I'm so-to-say over-experienced by sheer practice. On the other hand I think to remember to have read in Konrad Knopp's book, that divergent summation in the case of geometric series can be inserted in every analytical expression (I'll to check this possibly false memory when I've Knopp's book available again).      
But here is an example, where the identity fails obviously:
$$ e^{-1-2-4-8-16- \cdots }={1 \over e^1}{1 \over e^2}{1 \over e^4} \cdots \ne e^1 $$ 
How can I characterize the range of algebraic operations, where such (even much standard) analytic continuation is applicable and where not? (Other examples might be the insertion of $\zeta$-values at negative arguments in place of their sum/product-representations in algebraic formulae)     
[remark in the update 3]: analytic continuation needs some variable parameter with a possible value for which the expression is true/convergent. It proceeds in that that parameter gets changed as far the expression is analytic and convergent - and then analytic continuation is tried by further operations, coordinate change and shift of the range for the parameter. In the above formula such a variable parameter should be included, say the base parameter for the geometric series should be kept variable and for this the analytic continuation should then be attempted. This is kindly reflected in R.Israels answer
 
[update 2]: I'll add some context for this question from my comment to R. Israels's answer. It should shed much mor light on the intention of my question:
My question arose today when I re-read an older discussion of mine in the tetration-forum, where I didn't find an answer nor even a suitable direction for an answer (for my level of understanding at that time) Here is the link to the discussion where I had posed this in context with iteration-series and had already landed at that example of my current question: http://math.eretrandre.org/tetrationforum/showthread.php?tid=420 .

[update]:
I've done a quick view into the chapter "divergent series" in Knopp's monography (in german language). I see at least one formulation which I might have overgeneralized and not taken precisely enough. I'll paraphrase it here to show the root of my concern:      


(chap XIII, par. 261.) (...) "in a reasonable way" - this could also be interpreted, that we assign the sequence (s_n) in such a way a value s, that, wherever this sequence appears in a formula as a result of a computation, we should assign that value s always or at least generally to that result   (...)
  (par 262.) (...) Whether now, whereever this series $\sum (-1)^n $ occurs as a result of a computation, we should assign it the value $\frac 12$ - this cannot be decided without further consideration. With the representation $ {1 \over 1-x} = \sum x^n $ for $x=-1$ this however is surely the case. (...)        

It seems, I took that remarks too wide when I studied this chapter, and got too much unsensitive against the geometric series $\sum 2^k$ and its relatives...  possibly I should be more critical today even to Knopp's formulation, which seems a bit too vague in the light of my concern today.              
 A: Consider $f(z)=\dfrac{1}{1-z}$ as a function of one complex variable. Then $f$ is analytic near any $z\in\Bbb C$ except for $z=1,$ where it has a simple pole. The geometric series is the Taylor expansion of $f$ at $z=0,$ and because of the pole at $z=1,$ it has radius of convergence $1,$ meaning that $1+z+z^2+\cdots =\dfrac{1}{1-z}$ is valid for any complex number $|z|<1.$ For any other point $z_0$ in the complex plane, we can form a Taylor expansion $\sum_{n\ge 0}a_n(z-z_0)^n$ for $f$ with radius of convergence $R=|z_0-1|,$ but it will no longer be true that $a_n=1$ for all $n.$ The function $f$ is still well-defined and analytic at $z_0$ however.
A: I don't know what is your criterion for validity of the insertion.  Let's say $f$ is an analytic function on a domain $D$,  and the series $\sum_j g_j$ converges for $z$ in a domain  $W$ to a function $g$ that has an analytic continuation to a larger domain $U$, with $g(U) \subseteq D$.  Then $f(\sum_j g_j(z))$ for $z \in W$ has an analytic continuation to $f(g(z))$ on $U$.  Is that what you're thinking of?
In your example, with $f(z) = \exp(-z)$ and the series $\sum_{j=0}^\infty z^j$, $g(z) = 1/(1-z)$, it is indeed true that $$\exp\left(-\sum_{j=0}^\infty z^j\right) = \prod_{j=0}^\infty e^{-z^j}$$ has an analytic continuation to $\exp(-1/(1-z))$ on ${\mathbb C} \backslash \{1\}$, with value $e$ at $z = 2$.  However, I would avoid writing this as 
$$ \prod_{j=0}^\infty e^{-2^j} = e$$
EDIT: Note also that it is possible to have 
$f(\sum_{j=1}^N c_j z^j)$ converge uniformly on compact subsets of domain $A$ to an analytic function $g(z)$
and uniformly on compact subsets of domain $B$ to a different analytic function $h(z)$
where $g(z)$ and $h(z)$ are not analytic continuations of each other.  This will happen if $\sum_j c_j z^j$ has a finite nonzero radius of convergence and $f$ is analytic in a neighbourhood of $\infty$ and also analytic (and nonconstant) in a neighbourhood of $c_0$.  For example, with $f(z) = z/(1+z)$ and $\sum_{j=0}^\infty z^j$ we have
$$ \eqalign{\dfrac{\sum_{j=0}^N z^j}{1 + \sum_{j=0}^N z^j} &= \dfrac{z^{N+1}-1}{z^{N+1}-z+2}\cr
&\to 1 \ \text{for } |z| > 1\cr
& \to \dfrac{1}{2-z} \ \text{for } |z| < 1\cr}$$
A: You cannot analytically continue the sum in question beyond the unit disk, unless you omit $\{1\}$, in that your analytic continuation will not be valid at $x=1$. In your case, when you look at $e^{-1-2-4\ldots}$ you should in fact get 0 as you are raising an exponential to negative infinity.
