This is more like an extended comment to David Roberts' answer.
There (at least) two ways to see that
$\mathbb{A}^\infty\setminus\{0\}$ is $\mathbb{A}^1$-contractible. 


1. Using Morel's unstable connectivity theorem which can be found in his
book "$\mathbb{A}^1$-algebraic topology over a field", in the
chapter on $\mathbb{A}^1$-homotopy and $\mathbb{A}^1$-homology
sheaves. The result holds for the $\mathbb{A}^1$-homotopy theory over
an infinite perfect field, and states that any $n$-connected
simplicial presheaf is also $\mathbb{A}^1$-$n$-connected. 
It is claimed in Example 3.2.20 of [MV] and explained in Examples 2.11
of Dugger-Isaksen "Motivic cell structures" that there is an
$\mathbb{A}^1$-weak equivalence $\mathbb{A}^n\setminus\{0\}\cong
S^{n-1}\wedge\mathbb{G}_m^{\wedge n}$. In particular,
$\mathbb{A}^n\setminus\{0\}$ is simplicially $(n-2)$-connected, and by
Morel's unstable connectivity theorem also
$\mathbb{A}^1$-$(n-2)$-connected. As mentioned in S. Carnahan's
comment, taking the colimit of
$\mathbb{A}^n\setminus\{0\}$ via the obvious inclusions therefore
shows that $\mathbb{A}^\infty\setminus\{0\}$ is
$\mathbb{A}^1$-contractible. 

2. The alternative is the classical topological argument that
$S^\infty$ is contractible (see the MO-discussion
http://mathoverflow.net/questions/198/how-do-you-show-that-s-infty-is-contractible), made algebraic: 
the shift-by-1 map is 
$$
S:\mathbb{A}^\infty\setminus\{0\}\to\mathbb{A}^\infty\setminus\{0\}:
(x_1,x_2,\dots)\mapsto (0,x_1,x_2,\dots)
$$
which is homotopic to the identity via
$$
f_T:\mathbb{A}^\infty\setminus\{0\}\times\mathbb{A}^1\to
\mathbb{A}^\infty\setminus\{0\}:
(x_1,x_2,\dots)\mapsto (1-T)(x_1,x_2,\dots)+TS(x_1,x_2,\dots).
$$
This works since the straight line through $(x_1,x_2,\dots)$ and
$(0,x_1,\dots)$ does not go through $0$. 
Then the image of the shift map $S$ can be contracted to a point via 
$$
g_T:\mathbb{A}^\infty\setminus\{0\}\times\mathbb{A}^1\to
\mathbb{A}^\infty\setminus\{0\}:
(x_1,x_2,\dots)=(1-T)(0,x_1,x_2,\dots)+T(1,0,0,\dots). 
$$
This is just the same argument as in topology. In topology, the above
maps must be renormalized to induce maps on $S^\infty$, but that is
not necessary here. As the above maps are polynomial, they induce even
naive $\mathbb{A}^1$-homotopies from the identity
to the constant map, so $\mathbb{A}^\infty\setminus\{0\}$ is $\mathbb{A}^1$-contractible. 

The explicit map is then given as in David Robert's
answer. Formulated slightly differently, take
$\mathbb{A}^\infty\setminus\{0\}\to\mathbb{P}^\infty$ to be the
universal $\mathbb{G}_m$-bundle, and take the standard covering
$\mathcal{U}$ of $\mathbb{P}^\infty$ by the $\mathbb{A}^\infty$ with coordinates
$x_i\neq 0$, $i\in\mathbb{N}$. The universal $\mathbb{G}_m$-bundle
induces a map $\check{C}(\mathcal{U})\to B\mathbb{G}_m$. Here,
$\check{C}(\mathcal{U})$ is the Cech nerve of the cover
$\mathcal{U}$ which is simplicially equivalent to 
the action groupoid. The map itself is the $\mathbb{G}_m$-cocycle for 
the universal bundle in degree $1$, and can be extended to a map of
the whole Cech nerve since the cocycle condition is satisfied. The
diagram  $\mathbb{P}^\infty\leftarrow\check{C}(\mathcal{U})\to
B\mathbb{G}_m$ gives the required weak equivalence in 
$\mathbb{A}^1$-homotopy category. Note that the map itself is already
defined in the simplicial model category because
$\check{C}(\mathcal{U})\to \mathbb{P}^\infty$ is a simplicial weak
equivalence.