No, there is no such group.
By the structure theorem for finitely generated abelian groups,
$G$ can be written as a product of cyclic groups.
If any of these factors has odd order, then $G$ has an element of odd order,
so the product of the orders of all of its elements cannot be a power of 2.

We may thus consider only abelian $2$-groups hereafter.
For such a group $G$, the product of the orders of all of its elements
has the form $2^{k(G)}$ for some nonnegative integer $G$, and we must show
that it is impossible to achieve $k(G) = 2009$.
Again by the structure theorem, we may write
$$G \cong \prod_{i=1}^\infty (\ZZ/2^i \ZZ)^{e_i}$$
for some nonnegative integers $e_1,e_2,\dots$, all but finitely many of
which are $0$.

For any nonnegative integer $m$, the elements of $G$ of order at most $2^m$
form a subgroup isomorphic to
$$\prod_{i=1}^\infty (\ZZ/2^{\min\{i,m\}} \ZZ)^{e_i},$$
which has $2^{s_m}$ elements for $s_m = \sum_{i=1}^\infty \min\{i,m\} e_i$.
Hence
$$k(G) = \sum_{i=1}^\infty i(2^{s_i} - 2^{s_{i-1}}).$$
Since $s_1 \leq s_2 \leq \cdots$, $k(G)+1$ is always divisible by $2^{s_1}$.
In particular, $k(G) = 2009$ forces $s_1 \leq 1$.

However, the only cases where $s_1 \leq 1$ are where all of the $e_i$ are $0$,
in which case $k(G) = 0$, or where $e_i = 1$ for some $i$ and $e_j = 0$
for $j \neq i$, in which case $k(G) = (i-1)2^i + 1$.
The right side is a strictly increasing function
of $i$ which equals $1793$ for $i=8$ and $4097$ for $i=9$, so it can never equal
2009. This proves the claim.