= Solution
At fixed $r$, the leading equation is $(r^2f_0')'=0$. Its solution compatible with the eventual decaying far field and the boundary value at one is $f_0=1/r$. This decay implies $f=O(\epsilon)$ at distances $r=O(\epsilon^{-1})$. Balancing radial <derivatives> against the linear screening term identifies
$$
\boxed{\beta=1,\qquad x=\epsilon r.}
$$
The distant region is governed by the <modified Helmholtz equation>; its decaying homogeneous profile is $e^{-x}/x$.
For a <matched asymptotic expansion>, write the fixed-$r$ approximation as $f=f_0+\epsilon f_1+\epsilon^2f_2+\cdots$, allowing logarithms of $\epsilon$ in the coefficients. Successive equations are
$$
(r^2f_0')'=0,\qquad (r^2f_1')'=0,\qquad (r^2f_2')'=r+1.
$$
The boundary condition imposes $f_0(1)=1$ and $f_1(1)=f_2(1)=0$. Before matching, their integrated forms can be written
$$
f_0=\frac1r,\qquad f_1=C_1\left(1-\frac1r\right),\qquad f_2=\frac r2+\log r+D-\frac{D+1/2}{r}.
$$
In particular a pure power series with parameter-independent coefficients will be insufficient: the <logarithmic overlap creates a switchback term>.
In the distant region set $f=\epsilon F_0(x)+\epsilon^2F_1(x)+\cdots$. The scaled equation is $f_{xx}+2f_x/x-f=xf^3/\epsilon$, so
$$
F_0''+\frac2xF_0'-F_0=0,\qquad F_1''+\frac2xF_1'-F_1=\frac{e^{-3x}}{x^2}.
$$
Decay and leading matching give $F_0=e^{-x}/x$. The <radial modified Helmholtz equation> gives the supplied particular integral in terms of the <exponential integral>:
$$
F_1(x)=\alpha\frac{e^{-x}}x+\frac{e^{-x}E_1(2x)-e^xE_1(4x)}{2x}.
$$
As $x\downarrow0$, the <small-argument expansion of the exponential integral> yields
$$
F_1(x)=\frac{\alpha+\tfrac12\log2}{x}+\log x-\alpha+\tfrac32\log2+\gamma-1+O(x\log x),
$$
where $\gamma$ is the <Euler--Mascheroni constant>. Substitute $x=\epsilon r$ to compare the two expansions in $1\ll r\ll\epsilon^{-1}$:
$$
\epsilon F_0+\epsilon^2F_1=\frac1r-\epsilon+\epsilon^2\frac r2+\frac{\epsilon}{r}\left(\alpha+\tfrac12\log2\right)+\epsilon^2\left[\log r+\log\epsilon-\alpha+\tfrac32\log2+\gamma-1\right]+\cdots.
$$
The constant at order $\epsilon$ fixes $C_1=-1$. Its $1/r$ coefficient then fixes $\alpha+\tfrac12\log2=1$, and the constant at order $\epsilon^2$ fixes $D$. Hence
$$
\boxed{\alpha=1-\tfrac12\log2,\qquad D=\log\epsilon+2\log2+\gamma-2.}
$$
The required <inner expansion> at fixed $r$ is
$$
\boxed{f=\frac1r+\epsilon\left(\frac1r-1\right)+\epsilon^2\left[\frac{r-r^{-1}}2+\log r+\left(\log\epsilon+2\log2+\gamma-2\right)\left(1-\frac1r\right)\right]+o(\epsilon^2).}
$$
At fixed positive $x=\epsilon r$, the <outer expansion> is
$$
\boxed{f=\epsilon\frac{e^{-x}}x+\epsilon^2\left[\left(1-\tfrac12\log2\right)\frac{e^{-x}}x+\frac{e^{-x}E_1(2x)-e^xE_1(4x)}{2x}\right]+o(\epsilon^2).}
$$
Both display every term through the requested order, including the $\epsilon^2\log\epsilon$ <switchback term> in the fixed-$r$ region.
To form an <additive composite expansion>, subtract the common overlap from the sum of the inner and outer expressions. Their retained common part is
$$
f_{\rm overlap}=\frac1r+\epsilon\left(\frac1r-1\right)+\epsilon^2\left[\frac r2+\log r+D\right].
$$
Thus one composite is $\epsilon F_0(\epsilon r)+\epsilon^2F_1(\epsilon r)-\epsilon^2(D+1/2)/r$. The last term can be screened by multiplying it by $e^{-\epsilon r}$ without changing either retained expansion. This gives a useful exponentially decaying version:
$$
\boxed{f_{\rm comp}(r)=\frac{e^{-\epsilon r}}r\left[1-\epsilon^2(D+1/2)\right]+\epsilon^2F_1(\epsilon r).}
$$
Its boundary value is $1+o(\epsilon^2)$. If exact satisfaction of the boundary value is desired, use instead
$$
f_{\rm comp}^{\rm normalized}(r)=\left[1-\epsilon^2F_1(\epsilon)\right]\frac{e^{-\epsilon(r-1)}}r+\epsilon^2F_1(\epsilon r).
$$
The <small-argument expansion of the exponential integral> shows that this normalized composite has the same two retained expansions; it equals one at $r=1$ and tends to zero at infinity.
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