A fairly complete answer to this question appears in Woodin's "In search of Ultimate $L$" at the beginning of the section on Martin-Steel extender sequences. Woodin omits the proofs, so I'll fill in some of the details.

If $E$ is a short extender, then $L[E] = L[U]$ where $U$ is the normal measure of $E$. To see this, let $j : V\to M$ be the ultrapower of $V$ by $E$. Kunen's analysis shows that $j\restriction V_{\kappa+1}\cap L[U]$ coincides with the unique iterated ultrapower embedding $i : L[U]\to L[j_E(U)]$. (Every element of $V_{\kappa+1}\cap L[U]$ is $\Sigma_2$-definable in $L[U]$ from parameters in $\kappa\cup \Gamma$ where $\Gamma$ is the class of common fixed points of $i$ and $j$; since $i$ and $j$ are elementary and agree on these parameters, they agree on $V_{\kappa+1}\cap L[U]$.) Since $i$ is definable over $L[U]$, $E\cap L[U] = \{(a,X)\in L[U] : X\subseteq [\kappa]^{<\omega}, a\in i(X)\}$ is in $L[U]$. Since $L[E]$ is the minimum proper class model $N$ of ZFC such that $E\cap N \in N$, $L[E]\subseteq L[U]$. The reverse inclusion is obvious.

If $E$ is a long extender, then $L[E]$ can be larger than $L[U]$. For example, if $U$ and $W$ are measures on distinct measurable cardinals and $E$ is the extender of $j_W\circ j_U$, then $L[E] = L[U,W]$ has two measurable cardinals. Woodin's Theorem 4.20 asserts that if $E$ is a nice enough long extender (that is not "too long"), then $L[E]$ will not contain an inner model with a Woodin cardinal. This is probably pretty hard. Woodin also claims that dropping the niceness condition, consistently, one can code arbitrary sets into long extenders, so these do not have an inner model theory. There are still some open questions here, but these results together argue that inner model theory needs more than one extender.