Generating a finite group from elements in each conjugacy class

No, this is impossible. This is a standard lemma, but I'm finding it easier to give a proof than a reference: Let $G$ be your finite group. Suppose that $H$ were a proper subgroup, intersecting every conjugacy class of $G$. Then $G = \bigcup_{g \in G} g H g^{-1}$. If $g_1$ and $g_2$ are in the same coset of $G/H$, then $g_1 H g_1^{-1} = g_2 H g_2^{-1}$, so we can rewrite this union as $\bigcup_{g \in G/H} g H g^{-1}$. There are $|G|/|H|$ sets in this union, each of which has $|H|$ elements. So the only way they can cover $G$ is if they are disjoint. But they all contain the identity, a contradiction.

UPDATE: I found a reference. According to Serre, this result goes back to Jordan, in the 1870's.

It is impossible. As I mentioned in the comment to Richard Stanley's answer, you are looking for a finite group $G$ with a maximal subgroup $M$ such that $M$ intersects every conjugacy class. Then $G=\cup M^g$ is the union of $M$ and its conjugates, which is well-known to never happen.

Steve

The impossibility also follows from Jordan's lemma:

Let $G$ be a finite group which acts transitively on a set $\Omega$ with $|\Omega|:=n\geq 2$. Then there exists a $g\in G$ such that $\chi(g)=0$ where $\chi$ denotes the permutation character (put in simple terms this means that $g$ fixes no element of $\Omega$ ).

In fact with some additional work one can show that the proportion of elements $g\in G$ such that $\chi(g)=0$ is larger than or equal to $\frac{1}{n}$. So now let us see how Jordan's lemma implies that the answer to the OP's question is negative. So let $H$ be the group generated by $\{g_i\}$, a complete set of representatives of the conjugacy classes of $G$. Suppose that $H$ is a proper subgroup of $G$. Then we may look at the left action of $G$ on $G/H$. Since $|G/H|\geq 2$ and the action is transitive, it follows from Jordan's lemma that there exists a $x\in G$ such that for all $i$, $x g_i H\neq g_i H$. In other words, for each $g_i$ one has that $g_i^{-1}x g_i\notin H$ which in turn implies that for all $g\in G$ one has that $g^{-1}xg\notin H$; and therefore the conjugacy class of $x$ does not intersect $H$ which is absurd.

Note also that one gets the following corollary from the previous argument:

Let $H$ be a proper subgroup of $G$ then we may always find two distinct (linear) characters of $G$ that have the same restriction to $H$.

Indeed, by the previous argument there exists a conjugacy class $C$ of $G$ that does not intersect $H$. Let $D=G-C$ and define $f$ to be the class function which is equal to $0$ on $D$ and $1$ on $C$ and let $g$ be the class function which is equal to $1$ everywhere. Since $f$ and $g$ are (in a unique way) linear combinations of irreducible characters of $G$ and $f|H=g|H$ there must exist distinct irreducible characters $\psi_1$ and $\psi_2$ of $G$ which have the same restriction to $H$.