Subgroup conditions: inverse in the absence of an identity

This is a problem in Fraleigh's abstract algebra text.

$$F$$ contains the set of functions from $$\mathbb{R}$$ to $$\mathbb{R}$$, and $$\tilde{F}$$ is the subset of functions that are non-zero at any point in the domain. The question is whether the subset -- call it $$H$$ -- of such functions with the property $$f(0) = -1$$ are a group under multiplication.

We certainly do not have closure. Evaluating the produt of two functions, $$f$$, and $$g$$, at $$0$$ will give $$f(0) g(0) = (-1)(-1) = 1$$, which is not $$-1$$. We also certainly do not have an identity element, as such a function, $$e(x)$$, would be constant at $$1$$ for all $$x$$, meaning that $$e(0) \neq -1$$.

My question concerns inverses, though. It is my understanding that we cannot talk about inverses unless we have an identity element because it should be the case that an inverse, by definition, is such that for any $$a, b \in G$$, $$a*b = b*a = e$$ for some group $$G$$ and elements $$a, b$$ and identity $$e$$. However, if I take some $$f$$ in the group and define $$g = \frac{1}{f}$$, I get $$fg = gf = 1$$. and surely $$g(0) = \frac{1}{f(0)} = -1$$.

Thus, it seems that an inverse exists, but the identity function is not in the set. The set isn't closed under multiplication, so I certainly would not require that $$fg$$ be in the set, so it seems fine that it generates the identity. But this seems like a peculiar case, and it seems that I may have made a mistake somewhere.

Any help would be greatly appreciated.

In the group of integers under addition the set $$\{1, -1\}$$ isn't closed under the group operation, doesn't contain the identity, but is closed under the formation of additive inverses.