at.algebraic topology - unpointed brown representability theorem

Yes, Brown representability holds for such functors.
There are not really any material differences between this and the proof of Brown representability in the pointed case.

EDIT: My previous version of this was not rigorous enough. I was trying to be clever and get away with just simple cell attachments, which only work if you already know that the functor is represented by a space. Sorry for the delay in reworking, but this particular proof has enough details that it takes time to write up.

As you say, you begin by decomposing such functors so without loss of generality $F(pt)$ is a single point.

Start with $X_{-1}$ as a point. Assume you've inductively constructed an $(n-1)$-dimensional complex $X_{n-1}$ with an element $x_{n-1} in F(X_{n-1})$ so that, for all CW-inclusions $Z to Y$ of finite CW complexes with $Y$ formed by attaching a $k$-cell for $k < n$, the map
$$
[Y,X_{n-1}] to [Z,X_{n-1}] times_{F(Z)} F(Y)
$$is surjective.

Now, define a "problem" of dimension $n$ to be a CW-inclusion $Z to Y$ where $Y$ is a subspace of $mathbb{R}^infty$ formed by attaching a single $n$-cell to $Z$, together with an element of
$$
Map(Z,X_{n-1}) times_{F(Z)} F(Y).
$$
The fact that $Y$ has a fixed embedding in $mathbb{R}^infty$ means that there is a set of problems $S$, whose elements are tuples $(Z_s,Y_s,f_s,y_s)$ with $f_s$ a map $Z_s to X_{n-1}$ and $y_s$ is a compatible element of $F(Y)$.

Let $X_n$ be the pushout of the diagram
$$
X_{n-1} leftarrow coprod_{s in S} Z_s rightarrow coprod_{s in S} Y_s
$$
where the lefthand maps are defined by the maps $f_s$ and the righthand maps are the given $CW$-inclusions. This is a relative $CW$-inclusion formed by attaching a collection of $n$-cells; therefore, $X_n$ still has the extension property for relative cell inclusions of dimension less than $n$.

The space $X_n$ is homotopy equivalent to the homotopy pushout of the given diagram, which is formed by gluing together mapping cylinders. Specifically, $X_n$ is weakly equivalent to the space
$$
X_{n-1} times {0} cup (coprod_S Z_s times [0,1]) cup (coprod_S Y_s times {1})
$$
which decomposes into two CW-subcomplexes:
$$
A = X_{n-1} times {0} cup (coprod Z times [0,1/2])
$$
which deformation retracts to $X_{n-1}$, and
$$
B = (coprod Z_s times [0,1/2]) cup (coprod Y_s times {1})
$$
which deformation retracts to $coprod Y_s$ with intersection $A cap B cong coprod Z_s$. The Mayer-Vietoris property and the coproduct axiom then imply that there is an element $x_n in F(X_n)$ whose restriction to $A$ is $x_{n-1}$ and whose restriction to $B$ is $prod y_s$.

Taking colimits, you have a CW-complex $X$ with an element $x in F(X)$ (constructed using a mapping telescope + Mayer-Vietoris argument) so that, for all CW-inclusions $Z to Y$ obtained by attaching a single cell, the map
$$
[Y,X] to [Z,X] times_{F(Z)} F(Y)
$$ is surjective.

Now you need to show that for any finite CW-complex $K$, $[K,X] to F(X)$ is a bijection.

First, surjectivity is straightforward by induction on the skeleta of $K$. More specifically, for any $K$ with subcomplex $L$, element of $F(K)$, and map $L to X$ realizing the restriction to $F(L)$, you induct on the cells of $Ksetminus L$. Then, injectivity: if you have two elements $K to X$ with the same images in $F(K)$, you use the above-proven stronger surjectivity property to the inclusion $K times {0,1} to K times [0,1]$ to show that there is a homotopy between said maps.