Mathematica is able to compute these limits, the result is in terms of a partial derivative of the regularized hypergeometric function $_2\!\tilde{F}_1$. For $k=2$ I find

$$\lim_{k\rightarrow 2}x^{-k}I_k(x)=\frac{11}{3}-\gamma_{\rm{Euler}}+\ln x-\frac{1}{6x^3}\left(_2\!\tilde{F}_1^{(0,0,1,0)}(-3,1,-2;x)+\,_2\tilde{F}_1^{(1,0,0,0)}(-3,1,-2;x)\right).$$
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
$$\lim_{n\rightarrow k}x^{-n}I_n(x)=c_k-\gamma_{\rm{Euler}}+\ln x-\frac{1}{(k+1)!x^{k+1}}\left(_2\!\tilde{F}_1^{(0,0,1,0)}(-k-1,1,-k;x)+\,_2\tilde{F}_1^{(1,0,0,0)}(-k-1,1,-k;x)\right).$$
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 $k=2,3,4,5,6$.