Semisimple, connected Lie groups generated by unipotent elements. Let $G$ be a linear, semisimple Lie group with no compact factors.
The unipotent elements of $G$ are those that have only eigenvalue 1.
I've seen it asserted that $G$ is generated by its unipotent elements: see Exercise #2 $\S 4.5$ in Dave Witte Morris' book on Arithmetic Groups.
The hint in the book is that you consider the simple factors.
But even considering $SL(2, \mathbb{R})$, it is unclear to me why it is generated by its unipotent elements.
I am aware that any matrix in $G$ can be written as a product of commuting hyperbolic, elliptic, and unipotent element, but I am unsure of how you might generally express the hyperbolic and elliptic elements as unipotent elements.

I think I did the case $SL_2(\mathbb{R}):$
Let
$$A=\begin{pmatrix}
a&b\\c&d
\end{pmatrix} \in SL_2(\mathbb{R}).$$
Notice that
$$A\cdot\begin{pmatrix}
0&1\\-1&0
\end{pmatrix} = \begin{pmatrix}
-b&a\\-d&c
\end{pmatrix}, $$
and
$$\begin{pmatrix}
0&1\\-1&0
\end{pmatrix}\cdot A = \begin{pmatrix}
c&d\\-a&-b
\end{pmatrix},$$
and also
$$\begin{pmatrix}
0&1\\-1&0
\end{pmatrix} = \begin{pmatrix}
1&0\\-1&1
\end{pmatrix}\begin{pmatrix}
1&1\\0&1
\end{pmatrix}\begin{pmatrix}
1&0\\-1&1
\end{pmatrix}.$$
So, using these facts, you'll be able to reduce your problem to show that 
$$ A = \begin{pmatrix}
a&0\\0&1/a
\end{pmatrix}$$
is generated by unipotent elements.
But
$$ \begin{pmatrix}
a&0\\0&1/a
\end{pmatrix}
=
\begin{pmatrix}
1&0\\1/a-1&1
\end{pmatrix}
\begin{pmatrix}
1&1\\0&1
\end{pmatrix}
\begin{pmatrix}
1&0\\a-1&1
\end{pmatrix}
\begin{pmatrix}
1&-1/a\\0&1
\end{pmatrix}.$$
