The idea can be justified from a mathematical point of view. This justification uses the idea that, adding or subtracting a finite quantity, to or from an infinite one, cannot really make a difference to the infinity.
Let $r_a$ be the radial coordinate beyond which a real valued continuum wave function adopts its asymptotic form, and let $\Psi_{k,L}(r)$ be the radial part of such a wave function, in some central potential, so that $\Psi_{k,L,a}(r)$, is it's asymptotic form for any $r>r_a$.
For $\mathbf{k=k^\prime}$ , so that the first integral below is infinite
We have
\begin{align*}
\int _0^\infty \Psi_{k,L}(r) \Psi_{{k^\prime} ,L}(r) dr &= \int _0^{r_a} \Psi_{k,L}(r) \Psi_{{k^\prime },L}(r) dr + \int _{r_a}^\infty \Psi_{k,L,a}(r) \Psi_{{k^\prime} ,L,a}(r) dr \\
&=\int _{r_a}^\infty \Psi_{k,L,a}(r) \Psi_{{k^\prime} ,L,a}(r) dr \\
&=\int ^{r_a}_0 \Psi_{k,L,a}(r) \Psi_{{k^\prime} ,L,a}(r) dr + \int _{r_a}^\infty \Psi_{k,L,a}(r) \Psi_{{k^\prime} ,L,a}(r) dr\\
&=\int _0^\infty \Psi_{k,L,a}(r) \Psi_{{k^\prime} ,L,a}(r) dr
\end{align*}
The above "equations", should be approximately OK for $k$ approximately equal to $ k^\prime$ as well. So, how $ \int _0^\infty \Psi_{k,L}(r) \Psi_{{k^\prime} ,L}( r) dr $ behaves, as a multiple of a delta function $\delta(k-k^\prime)$ is determined by the asymptotic forms of $\Psi_{k,L}$ and $\Psi_{k^\prime ,L}$.
For some number $N_L$ we will have
\begin{equation*}
\int ^\infty_0 \Psi_{k,L,a}(r) \Psi_{{k^\prime} ,L,a}(r) dr = N_L \delta(k-k^\prime)
\end{equation*}
There is no mention of the form of $\Psi_{k,L}$ or $\Psi_{k^\prime ,L}$ close to the scattering centre in the above.
Hence, the normalisation of a continuum/scattering wave function is insensitive to the functions form near to the scattering centre.
