Denote the commutator of $a$ and $b$ by $a^{-1}b^{-1}ab = [a,b]$.
If $u$ is an element from the commutator subgroup, then $g^{-1}ug = u(u^{-1}g^{-1}ug) = u[u, g]$ .
Another approach: the commutator subgroup is defined to be the subgroup generated by the commutators, so every element of the commutator subgroup is of the form $$[a_1, b_1][a_2,b_2]\ldots[a_n, b_n].$$ It is enough to show that $g^{-1}[a,b]g$ is always in the commutator subgroup, because then
$$g^{-1}[a_1, b_1][a_2,b_2]\ldots[a_n, b_n]g = (g^{-1}[a_1, b_1]g)(g^{-1}[a_2,b_2]g)(g^{-1}\ldots g)(g^{-1}[a_n, b_n]g)$$
is a product of elements from the commutator subgroup. When $\phi$ is any homomorphism, we have $\phi([a,b]) = [\phi(a), \phi(b)]$. Since for any $g \in G$ the map $\phi$ defined by $\phi(x) = g^{-1}xg$ is a homomorphism, the result follows.
This is how I think of it:
Suppose $N$ is a subgroup containing every element of $G$ of the form: $aba^{-1}b^{-1}$ (these are called commutators).
If we knew $N$ was normal (back to that in a little bit) we could form the factor group $G/N$.
Now, in $G/N$ we have:
$(Na)(Nb)(Na)^{-1}(Nb)^{-1} = Naba^{-1}b^{-1} = N$
Which means that:
$(Na)(Nb) = (Nb)(Na)$, that is: $G/N$ is abelian.
OK, I promised I would talk about how we can be sure $N$ is normal. If we let $C$ be the set of all commutators, then the subgroup $\langle C\rangle$ of all finite products of commutators is certainly a subgroup containing all the commutators, and it's not hard to see it is the smallest such subgroup (we have to include at least all the commutators, and closure forces us to include all finite products).
Let's look at a conjugate of a commutator:
$g(aba^{-1}b^{-1})g^{-1} = ga(e)b(e)a^{-1}(e)b^{-1}(e)g^{-1}$
$= ga(g^{-1}g)b(g^{-1}g)a^{-1}(g^{-1}g)b^{-1}g^{-1}$
$= (gag^{-1})(gbg^{-1})(ga^{-1}g^{-1})(gb^{-1}g^{-1})$
$= (gag^{-1})(gbg^{-1})(gag^{-1})^{-1}(gbg^{-1})^{-1}$
In other words, a conjugate of a commutator is the commutator of the conjugates.
So the conjugate of any product of commutators is ALSO a product of commutators. This shows that $N = \langle C\rangle$ is normal.
A word about notation:
The expression $aba^{-1}b^{-1}$ is cumbersome to write, over and over again. The conjugates are also cumbersome to write over and over. So maths peoples have developed a "short-hand" by writing:
$aba^{-1}b^{-1} = [a,b]$
They (we?) also have a short-hand for writing:
$g^{-1}ag = a^g$ (with this notation we can say that: $(a^g)^h = a^{gh}$ and $(ab)^g = (a^g)(b^g)$, which is nifty).
In this shorthand, we can write:
$[a,b]^g = [a^g,b^g]$ which says what it took me 4 lines to write above (what I actually wrote was $[a,b]^{g^{-1}} = [a^{g^{-1}},b^{g^{-1}}]$ but I hope you get the picture).
With factor (quotient) groups, it is often common to suppress the "quotienting subgroup" and write:
$[a]$ or $\overline{a}$ instead of $Na$. In other words, we think of "factoring by $N$" as identically setting every element of $N$ to the identity (after all, $N = Ne = [e]$).
The "intuition" behind this is if $G$ were abelian, so that:
$ab = ba$
then:
$aba^{-1}b^{-1} = e$ for every $a,b$.
If we "set" all these elements equal to $e$ (which is what factoring does), that makes $G$ abelian (we might have to squish it a little bit). We can actually say a bit more:
Suppose $\phi: G \to G'$ is a group homomorphism such that $\phi(G)$ is abelian. Then $[G,G]$ (this is another way our commutator subgroup is represented) is contained in the kernel of $\phi$. This says that the commutator subgroup is always the SMALLEST subgroup that "kills" (sends to the identity of $G'$) all commutators, which is called the universal property of the commutator subgroup.
Because this construction is done "in the same way" for any group $G$, it is said to be categorical, and is also called the Abelianization of $G$. The group $G/[G,G]$ is the group that is "most like" $G$ but turned into an abelian group. If $G$ is already abelian, then all the commutators are just $e$, so $G/[G,G] = G/\{e\} = G$.
Best Answer
In general, the set of commutators does not equal the commutator subgroup, as pointed out above by several remarks. See also this paper of Kappe and Morse, where criteria are discussed for equality.
I. D. MacDonald provides counterexamples in [I. D. MacDonald, Commutators and Their Products, The American Mathematical Monthly Vol. 93, No. 6 (1986), pp. 440-444]. In particular, he proves by a simple counting argument the following theorem.
If $G$ is a finite group and $|G:Z(G)|^2<|G'|$, then $G'$ has elements which are not commutators.
Finally see also this StackExchange entry, which mentions another interesting paper of Marty Isaacs (also behind Jstor) with more counterexamples.