Error in Hungerford's algebra proof? Left id & inv = group Prop 1.3 in Hungerford's Algebra said that if $G$ is a semigroup and there exist a left identity and each element have a left inverse, then $G$ is a group. The proof (and in fact, even the proposition themselves) implicitly assumed that the left identity is unique, but I can't quite see how it could be proved. Is this a mistake, or is there a way to prove that?
EDIT: for definiteness, this is the exact statement as given:
Proposition 1.3. Let $G$ be a semigroup. Then $G$ is a group if and only if the following conditions hold:
(i) There exist an element $e\in G$ such that $ea=a$ for all $a\in G$ (left identity element);
(ii) For each $a\in G$, there exists an element $a^{-1}\in G$ such that $a^{-1}a=e$ (left inverse).
EDIT: in light of the discussion on the range of quantifier, the answer below showed that it is provable as long as the quantifier in (i) range all the way to include (ii). I still wish to ask whether we can prove it even if the quantifier only range within (i), ie. (ii) should be interpreted as: for each $a\in G$, there exists an element $a^{-1}\in G$ such that $a^{-1}a=e$ where $e$ is one of those left identities.
 A: Let's start by showing that left inverses are also right inverses. Take some element $g$ from the group, and denote a left identity by $e$. Then we know that 
$$ g^{-1} = eg^{-1} = (g^{-1}g)g^{-1} = g^{-1}(gg^{-1}).$$
Then this means that
$$ e = (g^{-1})^{-1}g^{-1} = (g^{-1})^{-1}\left(g^{-1}(gg^{-1})\right) = \left((g^{-1})^{-1}g^{-1}\right)gg^{-1} = egg^{-1} = gg^{-1},$$
so that the left inverse is also a right inverse. Notice that we did not use right identity or right inverse so far. With this, showing that the left identity is also a right identity is very easy.
$$ g = eg = (gg^{-1})g = g(g^{-1}g) = ge,$$
and so both the left identity and left inverses are actually also a right identity and right inverses. 
With this, you can carry out the standard proof that identities and inverses are unique. Suppose $e, e'$ are two identities. Then
$$e = ee' = e'.$$
Suppose $h, h'$ are both inverses to $g$. Then
$$ h = he = h(gh') = (hg)h' = eh' = h'.$$
So we get uniqueness and right-ness from having just a left identity and left inverse. $\diamondsuit$
A: I separate this answer, because it relies on an edit and a misreading of the original statement from Hungerford. 
But let us suppose that there is at least one left identity called $e$, and each element $g$ has a "loose left inverse" $g^{-1}$ such that $g^{-1}g$ is a left identity, which may or may not be $e$. 
Then we cannot conclude that $G$ is a group.
For instance, consider the two element semigroup with multiplication table
$$\begin{array}{l|ll}
G & e & g \\
\hline
e & e & g \\
g & e & g
\end{array}$$
In this case, every element is a left identity. For that matter, each element is a "loose left inverse" of both itself and the other element.
