After comparing Steven's original example with his new example, I believe I have an interesting generalization of both. Let $c \in \mathbb{C}$. Define the following three functions:
$$h(m)=\frac{1}{m-1+c},k(m) = \frac{1}{\Gamma(m+c)},$$
$$j(m)=\frac{h(m+1)}{k(m+1)}=\Gamma(m+c).$$
Let $\mathcal{H}(m,n),\mathcal{J}(m,n),\mathcal{K}(m,n)$ be the Hankel matrices corresponding to $h,j,k$, respectively. Let $H(m,n),J(m,n),K(m,n)$ be the respective determinants. Then the following holds:
$$H(m,n) =\pm J(m-1,n) K(m,n).$$
- When $c=1$, we recover the example from the original post.
- When $c=\frac{1}{2}$, we recover Steven's second example, from his answer to the post (up to some normalization).
Now I turn to prove the generalization.
Lemma 1: Let $V$ be an inner-product space. Let $\{v_i\}_{i=1}^{n},\{\tilde{v}_i\}_{i=1}^{n}$ be two pairs of sequences of linearly independent vectors in $V$, spanning the same subspace. Similarly, let $\{u_i\}_{i=1}^{n},\{\tilde{u}_i\}_{i=1}^{n}$ be two pairs of sequences of linearly independent vectors in $V$, spanning the same subspace.
There exist unique scalars $x_{i,j}, y_{i,j}$ such that $\tilde{v}_i = \sum x_{i,k} v_k$ and $\tilde{u}_i = \sum y_{i,k} u_k$.
Define the following 4 matrices:
- $A_{i,j} = \langle \tilde{v}_i, \tilde{u}_j \rangle$, $B_{i,j} = \langle v_i, u_j \rangle$
- $X_{i,j} = x_{i,j}, Y_{i,j} = y_{i,j}$
Then
$$A = X B Y^{T}.$$
Lemma 2: $\det(\binom{i+j+m}{i}_{0\le i,j \le n}) = 1$ for any $m \in \mathbb{C}$.
We work in the inner-product space $L^2([0,1])$.
Note that
$$\mathcal{H}(m,n)_{i,j} = \langle x^{i+m+c-2}, x^{j} \rangle$$
and that
$$\mathcal{K}(m,n)_{i,j} = \langle \frac{x^{i+m+c-2}}{\Gamma(i+m-1+c)}, \frac{(1-x)^j}{\Gamma(j+1)} \rangle.$$
We are in a situation where we may apply Lemma 1 with $A=\mathcal{H}(m,n)$ and $ B=\mathcal{K}(m,n)$, which yields
$$\mathcal{H}(m,n)= \text{Diag}((\Gamma(i+m-1+c))_{0\le i\le n}) \times \mathcal{K}(m,n) \times (\Gamma(i+1)\binom{j}{i}(-1)^i)_{0\le i,j \le n},$$
hence
$$(*) \frac{H(m,n)}{K(m,n)} = \det \text{Diag}((\Gamma(i+m-1+c))_{0\le i\le n}) \det (\Gamma(i+1)\binom{j}{i}(-1)^i)_{0\le i,j \le n}.$$
On the other hand,
$$\mathcal{J}(m-1,n) = (\Gamma(m-1+c+i+j))_{0 \le i,j \le n}) = \text{Diag}((\Gamma(i+m-1+c))_{0\le i\le n}) (\binom{m-2+i+j+c}{j})_{0 \le i,j \le n}) \text{Diag}(\Gamma(j+1)_{0\le j\le n}),$$
which, combined with Lemma 2, yields
$$(**) J(m-1,n) = \det \text{Diag}((\Gamma(i+m-1+c))_{0\le i\le n}) \det\text{Diag}(\Gamma(j+1)_{0\le j\le n}).$$
We finish by noting that
$$\det (\Gamma(i+1)\binom{j}{i}(-1)^i)_{0\le i,j \le n} = (-1)^{\binom{n+1}{2}} \prod_{i=0}^{n} \Gamma(i+1) = (-1)^{\binom{n+1}{2}} \det \text{Diag}(\Gamma(j+1)_{0\le j\le n}),$$
which implies that $(*)$ and $(**)$ coincide up to sign. $\blacksquare$
Best Answer
One thing you can say is that your determinant is the sum of determinants of the Cauchy matrices $C_S$ for subsets $S$ of $\{1,\ldots, n\}$, where $(C_S)_{i,j} = 1/(S_i + S_j - 1)$ for $1 \le i,j \le |S|$ (in the case $S = \emptyset$ we take the determinant to be $1$). This means $$ \det(I_n + H_n) = \sum_{S \subseteq \{1\ldots n\}} \frac{\prod_{1 \le i < j \le |S|} (S_i - S_j)^2}{\prod_{i,j=1}^{|S|} (S_i + S_j -1)}$$ For $n=1$ to $8$ your determinants are $$ 2,{\frac{29}{12}},{\frac{2927}{1080}},{\frac{659251}{224000}},{\frac{ 46508430817}{14817600000}},{\frac{616473989937916861}{ 186313420339200000}},{\frac{3577562384224548869428843}{ 1033954523962885324800000}},{\frac{1314142513507030576449489451528961} {365356847125734485878112256000000}} $$ I don't see an apparent pattern, nor does Maple's gfun package. The numerators and denominators don't seem to be in the OEIS.
EDIT: They are now. A295426 and A295427