Partial Fraction Expansion Algebra Help

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I hope someone can help. Given this equation

$$ F(s) = \frac{(1 - e^{-x})s^{-1}}{(1 - s^{-1})(1 - e^{-x}s^{-1})} $$

Apply a PFE

$$ = \frac{A_{1}}{1 - s^{-1}} + \frac{A_{2}}{1 - e^{-x}s^{-1}} $$

Then $$A_{1} = 1$$ $$A_{2} = -1$$

And

$$ F(s) = \frac{1}{1 - s^{-1}} - \frac{1}{1 - e^{-x}s^{-1}} $$

Same problem, but different solution.

$$ F(s) = \frac{(1 - e^{-x})s}{(s - 1)(s - e^{-x})} $$

PFE

$$ = \frac{A_{1}}{s - 1} + \frac{A_{2}}{s - e^{-x}} $$

$$A_{1} = 1$$

$$A_{2} = - e^{-x}$$

$$ F(s) = \frac{1}{s - 1} - \frac{e^{-x}}{s - e^{-x}} $$

Comparing that with the earlier answer the PFE coefficient for $A_{2}$ where we had $-1$ and then we have $-e^{-x}$ is puzzling. Which is correct? Why are the results so different? They should eventually give the same answer, so I am trying to see what steps I should take next. Or alternatively has a mistake been made somewhere?

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Both are correct, and not so different: $$\frac1{1-s^{-1}}=\frac s{s-1}=1+\frac1{s-1}$$ and $$\frac1{1-e^{-x}s^{-1}}=\frac s{s-e^{-x}}=1+\frac{e^{-x}}{s-e^{-x}}.$$

Only the second one is a true partial fraction decomposition w.r.t. the variable $s.$