Does $\mathbb C\mathbb P^\infty$ have a group structure?
I believe that the answer to the first question is yes, but I don't know about the second one.
It is well-known that ${\mathbb C}P^\infty$ is homeomorphic to the bar construction of the circle group $S^1$. You can see this from Milnor's model for the bar construction. The n-th skeleton of the bar construction is homeomorphic to the quotient $(S^1)^{*n+1}/S^1=S^{2n+1}/S^1={\mathbb C}P^n$.
It follows that ${\mathbb C}P^\infty$ has a structure of a topological abelian group. I don't know an explicit geometric description of this group structure.
{EDIT: As Tom pointed out, Milnor's model for $BG$ is, in most cases, homotopy equivalent but not homeomorphic to the bar construction, so this argument is inadequate. By inspecting low dimensional skeleta, it seems likely that in our case the two spaces are homeomorphic, by a non-obvious map. A later edit See also Ben Wieland's answer. }
It also is well-known that ${\mathbb C}P^n$ is homeomorphic to $Sp^n({\mathbb C}P^1)$, the n-fold symmetric product of ${\mathbb C}P^1$. You can construct a map $SP^n({\mathbb C}P^1)\to {\mathbb C}P^n$ as follows: let $[[u_1; v_1],\ldots, [u_n; v_n]]$ be a point of $SP^n({\mathbb C}P^1)$. Consider the the homogeneous polynomial of two variables $\prod_{i=1}^n(u_ix-v_iy)$. The coefficients of this polynomial define a point in ${\mathbb C}P^n$. The injectivity of this map follows from uniqueness of factorization of polynomials, and surjectivity follows from the fundamental theorem of algebra. Taking limit as $n$ goes to infinity, we obtain that ${\mathbb C}P^\infty$ is homeomorphic to $SP^\infty({\mathbb C}P^1)$ - the pointed free commutative monoid generated by ${\mathbb C}P^1$.
Note that ${\mathbb R}P^\infty$ is homeomorphic to the bar construction on the group ${\mathbb Z}/2$ by a similar argument (EDIT: With the same caveat as above). Thus ${\mathbb R}P^\infty$ also has an abelian group structure. However, as far as I know ${\mathbb R}P^\infty$ is not homeomorphic to a free commutative monoid.
I noticed that this question still has no accepted answer and all existing answers are rather long. It seems that the answer can be easily obtained using some results of infinite-dimensional topology, namely, the theory of manifolds modeled on the direct limit $\mathbb R^\infty$ of Euclidean spaces (see Chapter 5 of Sakai's book).
Two results of this theory will be important for our purposes:
Characterization Theorem 5.4.1. A topological space $X$ is is homeomorphic to an open subspace of $\mathbb R^\infty$ if and only if any embedding $f:B\to X$ of a closed subspace $B$ of a finite-dimensional compact metrizable space $A$ can be extended to an embedding $\bar f:A\to X$.
Classification Theorem 5.5.1. Two $\mathbb R^\infty$-manifolds are homeomorphic if and only if they are homotopically equivalent.
Now using the Characterization Theorem, it can be shown that $\mathbb{CP^\infty}$ is an $\mathbb R^\infty$-manifold.
Next, in his question Ben Wieland writes that $\mathbb{CP}^\infty$ is homotopically equivalent to the connected component $G_0$ of the free topological Abelian group $G$ over the sphere. Using the Characterization Theorem of Sakai once more, one can show (and this was done by Zarichnyi in 1982) that $G_0$ is an $\mathbb R^\infty$-manifold. Since $\mathbb{CP}^\infty$ and $G_0$ are two homotopically equivalent $\mathbb R^\infty$-manifolds, the Classification Theorem ensures that $\mathbb{CP}^\infty$ is homeomorphic to $G_0$ and hence has a compatible topogical group structure.
$\def\RR{\mathbb{R}}\def\CC{\mathbb{C}}\def\PP{\mathbb{P}}$This is a report of partial progress, not a full answer. Probably the most interesting thing, at the end, is that I show Segal's model of $B(\mathbb{Z}/2)$ really is $\mathbb{RP}^{\infty}$ and really does have an abelian group structure, but I wind up having to cite a pretty difficult paper to do it.
Let $G$ be a commutative metrizable group. According to these notes by Somnath Basu, Segal constructs $BG$ in the following manner. (I think that Basu has added a lot of insight of his own.) Let $EG_n$ be the set of all step functions $[0,1] \to G$ with at most $n$ discontinuities, where two step functions are equated if they disagree at finitely many points, metrized by (for example) $d_{EG_n}(\alpha, \beta) = \int_{[0,1]} d_G(\alpha(x), \beta(x)) dx$. For example, let $C^2$ be the cyclic group of order $2$. Then $EC^2_1$ consists of two closed line segments glued at their endpoints. Explicitly, the point $t$ in the first copy of $[0,1]$ corresponds to $$f_t(x) := \begin{cases} 0 & x<t \\ 1 & x>t \end{cases}$$ and the $t$ in the second copy corresponds to $$g_t(x) := \begin{cases} 1 & x<t \\ 0 & x>t. \end{cases}$$ The end points are glued because $f_0=g_1$ and $f_1=g_0$.
Similarly, $ES^1_1$ is $S^1 \times [0,1]$ with $\left( S^1 \times \{ 0,1 \} \right) \cup \left( \{0 \} \times [0,1] \right)$ contracted to a point, which is clearly $S^2$.
Segal defines $EG$ as the direct limit $\bigcup EG_n$, and $BG_n$ and $BG$ as the obvious quotients $EG_n/G$ and $EG/G$. Pointwise multiplication gives a map $EG_m \times EG_n \to EG_{m+n}$ and, if $G$ is commutative, it descends to the quotient $BG_m \times BG_n \to BG_{m+n}$. On the infinite unions $BG$ and $EG$, we get groups.
Remark: There is a natural surjection $F(BG_1) \to BG$, since $BG$ is generated by $BG_1$ as a group, but the restriction to the identity component of $F(BG_1)$ isn't an isomorphism. Let $$\phi(t,g)(x) = \begin{cases} 1 & x < t \\ g & x>t \end{cases}.$$ Then $\phi(t_1,g_1) + \phi(t_1,-g_1) - \phi(t_2, g_2) - \phi(t_2, -g_2)$ is in the connected component of the identity and in the kernel of the map. So this doesn't seem like a good way to approach the free group issue.
So we would like to know, for this particular construction, are $(EC^2)_n$, $(BC^2)_n$, $(ES^1)_n$ and $(BS^1)_n$ homeomorphic to $S^n$, $\mathbb{RP}^n$, $S^{2n+1}$ and $\mathbb{CP}^n$ respectively.
Set $EG^{\circ}_n := EG_n \setminus EG_{n-1}$ and $BG^{\circ}_n := BG_n \setminus BG_{n-1}$. Then $EG_n^{\circ}$ clearly isomorphic to $G \times (G \setminus \{ 0 \})^{n-1} \times \{ 0 < x_1 < x_2 < \cdots < x_n < 1 \} \cong G \times (G \setminus \{ 0 \})^{n-1} \times \RR^n$, and $BG_n^{\circ} \cong (G \setminus \{ 0 \})^{n-1} \times \RR^{n-1}$.
We see that $(EC^2)_n^{\circ} = \RR^n \sqcup \RR^n$, so $EC_n$ has a cell structure with two cells of each dimension. If we knew it was a regular cell decomposition, then we would know $(EC^2)_n \cong S^n$. Similarly, $(BS^1)_n^{\circ} \cong \RR^n \times \RR^n \cong \CC^n$, but it seems hard to understand the attachment maps.
I have seen $(EC^2)_n$ before! Consider $n=3$ first. $(EC^2)_3$ is divided into two maximal cells. Let $\Delta_3 = \{ (x_1, x_2, x_3) \in \RR_{\geq 0}^3 : \sum x_i = 1 \}$. Each maximal cell is of the form $(z,0,1-z) \cong (z',0,1-z')$ for any $z$, $z'$. We can explicitly build the quotient map $\Delta^3 \to (EC^2)^3$ as a map from $\Delta^3$ to $3 \times 3$ matrices by $$ (x,y,z) \mapsto \begin{pmatrix} 1 & x & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & y \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & z & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} . $$ The image is the matrices $\left( \begin{smallmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{smallmatrix} \right)$ obeying $a$, $b$, $c$, $\det \left( \begin{smallmatrix} a & b \\ 1 & c \end{smallmatrix} \right) \geq 1$, $a+c=1$. Think of this as related to the reduced word $s_1 s_2 s_1$ in the Weyl group of $GL_3$. Fomin and Shapiro build similar maps $\Delta^n \to U$ where $U$ is the unipotent radical of any reductive group $G$, corresponding to reduced words of length $n$ in the Weyl group of $G$. They build a corresponding cell complex and conjecture it is regular.
Hersh proved Fomin and Shapiro's conjecture! Theorem 6.13 of her paper shows that the cell complex for $(EC^2)_n$ is regular. (The CW complex she is discussing the closure of one of the two cells in our cellular structure.) I would love to see someone write out a simpler proof of this result, though. I went through Hersh's proof line by line and convinced myself it worked, but I never felt like I understood it conceptually.
It is also worth noting there is a different set of models: I'll call them $\widehat{EG}_n$ and $\widehat{BG}_n$ which are clearly isomorphic to $\CC \PP^n$ and $\RR \PP^n$.
A point of $\widehat{EG}_n$ is a pair $( (x_1, x_2, \ldots, x_n), f )$ where $0 \leq x_1 \leq x_2 \leq \cdots \leq x_n \leq 1$ and $f$ is a function in $EG_n$ whose discontinuity set is contained in $(x_1, x_2, \ldots, x_n)$. For example, $(EC^2)_1$ is made of four line segments glued into a square: Two of them are $\{ (t,f_t) \}$ and $\{ (t, g_t) \}$ for $f_t$ and $g_t$ as above; the other two are $\{ (t,0) \}$ and $\{ (t,1) \}$, where $0$ and $1$ are the constant functions. Again, $\widehat{BG}_n = \widehat{EG}_n/G$.
There are obvious maps $\widehat{EG}_n \to EG_n$ and $\widehat{BG}_n \to BG_n$. If $f$ has $k$ discontinuities, its preimage in $\widehat{EG}_n$ is a simplex of dimension $n-k$ (just specify the locations for the $n-k$ points not forced by $f$).
It is also easy to see that $\widehat{ES^1}_n \cong S^{2n+1}$ (and likewise for the other three cases). Given $((x_1, \ldots, x_n), f)$, formally set $x_0=0$ and $x_{n+1}=1$, and let $\theta_k$ be the value of $f$ on $(x_{k-1}, x_k)$. Set $r_k = \sqrt{x_{k}-x_{k-1}}$. Identify $((x_1, \ldots, x_n), f)$ with $$(r_1 \cos \theta_1, r_1 \sin \theta_1, r_2 \cos \theta_2, r_2 \sin \theta_2, \ldots, r_n \cos \theta_n, r_n \sin \theta_n).$$ Notice that $\theta_k$ is undefined precisely when $r_k=0$, so this makes sense, and it is easy to check that it gives a homeomorphism $\widehat{ES^1}_n \cong S^{2n+1}$. But there doesn't seem to be a group structure on $\widehat{BG}$.