Assume the powerseries of f(x) as $f(x)=1+a x+bx^2+cx^3+dx^4...$ where we deal with the explication of the problem just up to the coefficient $d$. Then I have a description of the $p$'th power of $f$ with $p$ as a variable, which may be best displayed with the help of a table.
Remark: I adopted a binomial notation, where in my original derivation is a pochammer-symbol. So for $p*(p-1)*(p-2)/3!$ in the original I write $\binom p3$ for shortness here. Thus for negative $p$ the binomial-expressions must respectively be interpreted!
Here is the beginning of the table for
$f(x)^p$ with positive
$p$ :
$ \small \begin{array} {rllll} f(x)^p=x^0* & \binom{p}{0}*(1)\\ +x^1*[& \binom{p}{1}*(1a) ] \\ +x^2*[& \binom{p}{1}*(1b)&+ \binom{p}{2}* (1a^2) ] \\ +x^3*[& \binom{p}{1}*(1c)&+ \binom{p}{2}* (2ab)&+ \binom{p}{3}* (1a^3)] \\ +x^4*[& \binom{p}{1}*(1d)&+ \binom{p}{2}* (2ac+1b^2)&+ \binom{p}{3}* (3a^2b)&+ \binom{p}{4}* &(a^4)] \\ +\ldots&\ldots \\ \end{array} $
If we insert
$p=1$ for
$f(x)^1$ we get the obvious correct result
$ \qquad \begin{array} {rllll} f(x)^1=x^0* & 1*(1)\\ +x^1*[& 1*(1a) ] \\ +x^2*[& 1*(1b) ] \\ +x^3*[& 1*(1c) ] \\ +x^4*[& 1*(1d) ] \\ +\ldots&\ldots \\ \end{array} $
If the above table is indeed a complete description for all
$p$, then using
$p=-1$ for the multiplicative inverse
$\frac{1}{f(x)}$ , this would give:
$ \qquad \begin{array} {rllll} f(x)^{-1}=x^0* & 1*(1)\\ +x^1*[& -1*(1a) ] \\ +x^2*[& -1*(1b)&+1*(1a^2) ] \\ +x^3*[& -1*(1c)&+1*(2ab)&-1*(1a^3)] \\ +x^4*[& -1*(1d)&+1*(2ac+1b^2)&-1*(3a^2b)&+ 1*(a^4)] \\ +\ldots&\ldots \\ \end{array} $
Then to have nonnegative coefficients in
$f(x)$ as well as in
$f(x)^{-1}$ means to have
1)
$a=0$ because of
$x^1$ 2) then
$b=0$ because of
$x^2$ where also
$a^2=0$ 3) and so on.
So if the above description (which was derived from positive powers $f(x)^p$ only) is sufficient also for the negative powers, then this should be acceptable as a proof, that only the constant function $f(x)=1$ has nonnegative coefficients in the formal powerseries of $f(x)$ and $f(x)^{-1}$ simultanously.
[Update]: I apply the correction according to Robert Israel, and get the following result.
Because in &f(x)& all coefficients are nonnegative, I can, without loss of generality rewrite
$\qquad f(x)=1+ax+b^2x^2+c^3x^3+d^4x^4+...$
Then the table for $f(x)^{-1}$ looks like
$ \qquad \begin{array} {rllll} f(x)^{-1}=x^0* & 1*(1)\\ +x^1*[& -1*(1a) ] \\ +x^2*[& -1*(1b^2)&+1*(1a^2) ] \\ +x^3*[& -1*(1c^3)&+1*(2ab^2)&-1*(1a^3)] \\ +x^4*[& -1*(1d^4)&+1*(2ac^3+1b^4)&-1*(3a^2b^2)&+ 1*(a^4)] \\ +\ldots&\ldots \\ \end{array} $
and from
1) $x^2$ follows that the minimal possible b equals a, so $b\ge a$
2) $x^3$ follows that the minimal possible c equals a, so $c\ge b\ge a$
3) and so on
So we have $ 0 < a \le b \le c \le d \le \ldots $
If all coefficients after a take their minimum, then $0 then the coefficients at all $x^k, k>1$ simplify to binomial coefficients whose sum is zero at the same power $a^k$, so the complete coefficient is zero as well. Thus $f(x)^{-1} = 1-ax $ , then $f(x)={1 \over 1-ax}$ and we have nonnegative coefficients in $f(x)$ and nonpositive in $f(x)^{-1}$ after the constant.
If we take $b>a$ at $x^2$ then at $x^3$ follows, that $c^3=b^3$ suffices to give a negative value at $x^3$ in $f(x)^{-1}$ .
Analoguos reasoning inherits to all following coefficients $0 leading to $f(x)$ has positive coefficients only and $f(x)^{-1}$ has negative coefficients after the constant, but I didn't look yet at the coefficients $d,e, \ldots$ to arrive at Robert Israel's solution.