$\newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\R}{\mathbb{R}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}}$
Is there an easy way to see why this should generalize? Yes.
Indeed, this can be extended to any number of blocks. Say a matrix $M$ consists of $q\times q$ blocks, with its $ij$-block given by the formula \begin{equation} a_{ij}1_{r_i\times r_j}+b_i\de_{ij}I_{r_i}, \end{equation} where $i,j=1,\dots,q$ and $\de_{ij}$ is the Kronecker symbol. The $ij$-block of the inverse matrix $M^{-1}$ has the same form: \begin{equation} c_{ij}1_{r_i\times r_j}+d_i\de_{ij}I_{r_i} \end{equation} for some real $c_{ij}$ and $d_i$.
Noting that $1_{r\times s}1_{s\times t}=s1_{r\times t}$, we see that the $ij$-block of the identity matrix $MM^{-1}$ is
\begin{equation}
\sum_k (a_{ik}1_{r_i\times r_k}+b_i\de_{ik}I_{r_i})(c_{kj}1_{r_k\times r_j}+d_k\de_{kj}I_{r_k})
=u_{ij}1_{r_i\times r_j}+b_id_i\de_{ij}I_{r_i},
\end{equation}
where
\begin{equation}
u_{ij}:=a_{ij}d_j+\sum_k(a_{ik}r_k+\de_{ik}b_k)c_{kj}.
\end{equation}
That is,
\begin{equation}
D_d=D_{1/b},\quad AD_b+(AD_r+D_b)C=0,
\end{equation}
where $D_v$ stands for the diagonal matrix with the coordinates of the vector $v=(v_1,\dots,v_q)$ on the diagonal, $d:=(d_1,\dots,d_q)$, $b:=(b_1,\dots,b_q)$, $1/b:=(1/b_1,\dots,1/b_q)$, $r:=(r_1,\dots,r_q)$, $A:=(a_{ij})$, and
\begin{multline*}
C:=(c_{ij})=-(AD_r+D_b)^{-1}AD_b
=-(I_q+D_r^{-1}A^{-1}D_b)^{-1}(AD_r)^{-1}AD_{1/b} \\
=-(I_q+D_r^{-1}A^{-1}D_b)^{-1}D_{1/(br)}\sim-D_{1/(br)}
\end{multline*}
if, for instance, $A^{-1}$ exists and $\min_i|r_i|\to\infty$, where $br:=(b_1r_1,\dots,b_qr_q)$. Thus, the $ij$-block $c_{ij}1_{r_i\times r_j}+d_i\de_{ij}I_{r_i}$ of the inverse matrix $M^{-1}$ behaves as
\begin{equation}
\Big(\frac1{b_i}\,I_{r_i}-\frac1{b_ir_i}\,1_{r_i\times r_i}\Big)\de_{ij}I_{r_i},
\end{equation}\begin{equation}
\Big(\frac1{b_i}\,I_{r_i}-\frac1{b_ir_i}\,1_{r_i\times r_i}\Big)\de_{ij},
\end{equation}
as desired.