Let $P(a,b,c)$ be the $n\times n$ matrix where
$$
P(a,b,c)_{ij} = \begin{cases}
a & \text{ if } i,j < n \\
b & \text{ if } i < n \text{ and } j = n \\
b & \text{ if } i = n \text{ and } j < n \\
c & \text{ if } i = j = n.
\end{cases}
$$

If I understand correctly, you want the square root of $I+P(r,r,0)$. You can check that
$$ (I+P(a,b,c))^2 = I + P((n-1)a^2+b^2+2a,(n-1)ab+bc+2b,(n-1)b^2+c^2+2c). $$
Thus, you just need to solve
\begin{align*}
(n-1)a^2+b^2+2a &= r \\\\
(n-1)ab+bc+2b &= r \\\\
(n-1)b^2+c^2+2c &= 0.
\end{align*}
Maple tells me that if we put
\begin{align*}
p &= \sqrt{1+(n-1)(r-r^2)} \\\\
q &= \sqrt{\frac{2p+2+(n-1)r}{(n-1)(n+3)}}
\end{align*}
then
\begin{align*}
a &= q + \frac{q-r-pq}{(n-1)r} \\\\
b &= q \\\\
c &= \frac{q-r-pq}{r}.
\end{align*}

UPDATE:

The square root of $I+P(r,r,r-1)$ can be done similarly. If we put
\begin{align*}
s &= (1 - 2\sqrt{r} + nr)^{-1/2} \\\\
a &= rs + \frac{s - \sqrt{r}s - 1}{n-1} \\\\
b &= rs \\\\
c &= rs - \sqrt{r}s - 1
\end{align*}
then
$$ \sqrt{I + P(r,r,r-1)} = 1 + P(a,b,c). $$