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TL;DR

$$\lim_{n\rightarrow k}x^{-n}I_n(x)=\frac{x-1}{a_k x^{k+1}}\left(\sum_{p=0}^k b_{p,k} x^p\right) -2 \,\text{arctanh}\,(1-2 x)$$

with $a_k$ in OEIS:A003418 and $b_{p,k}$, $p=0,1,2\ldots k$, the rows of the triangle in OEIS:A027446.

In particular,

$$\lim_{k\rightarrow 2}x^{-k}I_k(x)=\frac{(x-1) (11 x^2+5 x+2)}{6 x^3}-2 \,\text{arctanh}\,(1-2 x)$$

$\newcommand\HGF{_2\!\tilde{F}_1}$Mathematica is able to compute these limits, the result is in terms of a partial derivative of the regularized hypergeometric function $\HGF$. For $k=2$ I find $$\lim_{k\rightarrow 2}x^{-k}I_k(x)=\frac{11}{3}-\gamma_{\text{Euler}}+\ln x-\frac{1}{6x^3}\left({\HGF^{(0,0,1,0)}}(-3,1,-2;x)+{\HGF^{(1,0,0,0)}}(-3,1,-2;x)\right).$$ The superscript notation is Mathematica's way of indicating which variable to differentiate. As a test, for $x=1/2$ the right-hand-side evaluates to $-4.833333\cdots=-29/6$, which agrees with a numerical evaluation of $\lim_{k\rightarrow 2}[\beta_x( -1 - k, 0) + H_{-2 - k}]$.

The corresponding formula for integer $k\geq 2$ is $$\lim_{n\rightarrow k}x^{-n}I_n(x)=c_k-\gamma_{\text{Euler}}+\ln x-\frac{1}{(k+1)!x^{k+1}}\left({\HGF^{(0,0,1,0)}}(-k-1,1,-k;x)+ {\HGF^{(1,0,0,0)}}(-k-1,1,-k;x)\right).$$ The fraction $c_k$ is twice the constant term in an expansion of $H_{-2-n}$ around $n=k$. I don't have a closed-form expression for $c_k$, the first few values are $11/3, 25/6, 137/30, 49/10, 363/70$, for $k=2,3,4,5,6$.

Update: As pointed out by Peter Taylor: $c_k=2H_{k+1}$.

Mathematica code.

 k=6;
 Series[Gamma[a]x^a Hypergeometric2F1Regularized[a,1-b,a+1,x],{a,-1-k,0}]/.b->0//Normal// FullSimplify;
 %/.a->-1-x;
 Series[HarmonicNumber[-2-x],{x,k,0}]//Normal;
 %+%%//FullSimplify

The asympotics near $x=0$ and $x=1$ is $$\lim_{n\rightarrow k}x^{-n}I_n(x)\rightarrow\begin{cases} -\frac{1}{(k+1)x^{k+1}}&\text{for}\;\;x\rightarrow 0\\ -\ln(1-x)&\text{for}\;\;x\rightarrow 1 \end{cases}$$

Here is a plot of $\lim_{n\rightarrow k}x^{-n}I_n(x)$ for $k=2$.

Special cases: