Why does the map $BG\to A(*)$  fail to split? There is a map $BG \to A(\ast)$ where $BG$ classifies stable spherical fibrations and $A(\ast)$ is
Waldhausen's algebraic $K$-theory of a point. The map is induced by applying Quillen's plus construction to the inclusion 
$$
BGL_1(S^0) \to BGL_\infty(S^0)
$$ 
where $BGL_1(S^0)$ is $BG$. Here $BGL_\infty(S^0)$ can be defined as the homotopy colimit 
over $k$ and $n$ of $BG(\vee_k S^n)$, where $G(\vee_k S^n)$ is the topological monoid of homotopy automorphisms of a $k$-fold wedge of $n$-spheres. (Note: $A(\ast) = \Bbb Z \times BGL_\infty(S^0)^+$.)
 Question 1:   I've heard it mentioned that there can be  no  retraction $A(*) \to BG$ to the map $BG \to A(\ast)$.  Can someone please explain to me why there is no such splitting, and if possible, give a reference?
More generally,
 Question 2:  If $R$ is a structured ring spectrum, is there a reasonable set of conditions which guarantees that the map  $BGL_1(R)\to K(R)$ admits a retraction?  
(A related question is under what conditions does an $R$-module map
$f: R^n \to R^n$ which is a weak homotopy equivalence 
admit a determinant $\text{det}(f) \in GL_1(R)$).
 A: Question 1: There are several arguments.
In degree 2 there is a reference: the proof of corollary 3.7 of Waldhausen's "Algebraic K-theory of spaces, a manifold approach".  See http://www.math.uni-bielefeld.de/~fw/ for a copy.  Consider the maps $BG \to A(\ast) \to K(Z)$ and apply $\pi_2$.  Here $\pi_2 BG = Z/2$, the composite is zero (restrict to $BSG$) and $\pi_2 A(\ast) \to K_2(Z)$ is an isomorphism.  Hence the first homomorphism cannot be injective.
In degree 3 there is another argument.  Right composition with the unstable map $\eta : S^3 \to S^2$ takes the generator of $\pi_2(BO)$ to zero in $\pi_3(BO) = 0$, so by naturality with respect to $BO \to BG$, composition with $\eta$ takes the generator of $\pi_2(BG) = Z/2$ to zero in $\pi_3(BG)$.  On the other hand, composition with $\eta$ takes the generator of $\pi_2 Q(S^0)$ to a nonzero element in $\pi_3 Q(S^0)$, so by naturality with respect to $Q(S^0) \to A(\ast)$, composition with $\eta$ is nonzero from $\pi_2 A(*) = Z/2$ to $\pi_3 A(\ast)$. Therefore $BG$ cannot split off $A(*)$.
In degree 4 there is your argument.
Question 2: So far, I only know that $BGL_1(R) \to K(R)$ admits a retraction if $R$ is
the realization of a commutative simplicial ring.  There is no retraction for $R = ku$ by Corollary 2.3 of "Divisibility of the Dirac magnetic monopole as a two-vector bundle over the three-sphere" by Ch. Ausoni, B. I. Dundas and J. Rognes, Documenta Mathematica 13 (2008) 795-801, and ku is already pretty close to a commutative simplicial ring.
A: I just realized how the argument for Question 1 might go (I hope this isn't self-indulgence on my part):
The composite $BO \to BG \to A(*)$ factors through $Q(S^0)$.  So it suffices to show there is no map $Q(S^0) \to BG$ such that the composite $BO \to Q(S^0) \to BG$ is homotopic to the usual map.  If this map existed then the composite $\pi_4(BO) \to \pi_4^{\text{st}}(S^0) \to \pi_4(BG)$ would be the map $\Bbb Z \to \Bbb Z_{24}$ which is the J-homomorphism in this dimension, so it's non-trivial. 
However, we get a contradiction with this since $\pi^{\text{st}}_4(S^0)$ is trivial. 
Is this the usual proof?
