If $H$ is closed under multiplication and contains the inverse of each of its elements, then it contains also the identity as long as $H$ is non-empty.
Indeed
$$
H\neq\emptyset\Rightarrow
\exists h\in H\Rightarrow
\{h,h^{-1}\}\subset H\Rightarrow
1=hh^{-1}\in H.
$$
Here are two reasons the group generated by $U$ is defined as the set of finite products, which I will denote by $\Pi(U)$:
1.) The group generated by $U$ is usually taken to be the smallest group containing $U$; it is evident the set $\Pi(U)$ satisfies this criterion, since it is clearly closed under the group operation (finite products of finite products of elements of $U$ are, after all, themselves finite products of elements of $U$) and the taking of inverses, and contains the identity element $e$ since
$e = xx^{-1}, \; x \in U; \tag 1$
thus $\Pi(U)$ is a group; and any group containing $U$ must contain $\Pi(U)$ if it is to be closed under the group operation and inversation. Indeed, $\Pi(U)$ is often though of as he intersection of all groups containing $U$; in this sense it is the smallest group containing $U$.
2.) We really can't define infinite products of elements of $U$ anyway, in a purely algebraic sense; to do so generally requires some notion of $convergence$ of a sequence of products such as
$x_1x_2, x_1x_2x_3, x_1x_2x_3x_4, \ldots; \tag 2$
but convergence lies in the realm of topology, so we would have to adopt some appropriate topological structure to give meaning to such infinite products.
Well, there are two of my main reasons for accepting the definition of the group generated by $U$ as $\Pi(U)$. The comment stream attached to the question itself contains more useful insights, cf. the remarks of ThorWitch and Captain Lama.
Best Answer
For any element $x$ in a finite group $G$, $x^{|x|}=e$ by definition, so $e$ will automatically be in any generating set. Here $|x|$ denotes the order of $x$ in $G$.