Let's see how we can guess why $\theta_n$ satisfies that particular kind of equation, and why $u_n$ should be expected to have a limit. First write $$
r_n^{2n+1} - \left(1+\frac 1n\right)r_n^{2n} - \left(1+\frac 1n\right)r_n+1 = 0.
$$
Now, substitute $r_n = 1+\frac{\theta_n}{n}$ in this equation : $$
\left(1+\frac{\theta_n}{n}\right)^{2n+1}- \left(1+\frac 1n\right)\left(1+\frac {\theta_n}n\right)^{2n} -\left(1+\frac 1n\right)\left(1+\frac {\theta_n}n\right) + 1=0$$
Let's rearrange this by factorizing out $\left(1+\frac{\theta_n}{n}\right)^{2n}$ from the first two terms and multiplying out the third term. Simple algebra leads to : $$
\left(1+\frac{\theta_n}{n}\right)^{2n}\left(\frac{\theta_n-1}{n}\right) - \left(\frac{1+\theta_n}{n} + \frac{\theta_n}{n^2}\right) = 0
$$
Let's take the second term to the RHS and then divide both sides by it (which is legal as it is a positive term, since $\theta_n>0$ as $r_n > 1$). This leads to $$
\frac{\left(1+\frac{\theta_n}{n}\right)^{2n}\left(\frac{\theta_n-1}{n}\right)}{\left(\frac{1+\theta_n}{n} + \frac{\theta_n}{n^2}\right)} = 1
$$
A little bit of simplification with the $n$s yields $$
\frac{\left(1+\frac{\theta_n}{n}\right)^{2n}\left(\theta_n-1\right)}{\left(1+\theta_n + \frac{\theta_n}{n}\right)} = 1
$$
Time to apply the heuristics : we expect the approximations $\left(1+\frac{\theta_n}{n}\right)^{2n} \approx e^{2 \theta_n}$, and $\left(1+\theta_n + \frac{\theta_n}{n}\right) \approx 1+\theta_n$. Let's separate these out as appropriate ratios.
$$
\frac{\left(1+\frac{\theta_n}{n}\right)^{2n}}{e^{2 \theta_n}} \frac{\left(1+\theta_n\right)}{\left(1+\theta_n + \frac{\theta_n}{n}\right)}\frac{e^{2 \theta_n}\left(\theta_n-1\right)}{\left(1+\theta_n\right)} = 1
$$
Recognizing the third term as $u_n$, we can write $$
u_n = \frac{e^{2 \theta_n}}{\left(1+\frac{\theta_n}{n}\right)^{2n}} \frac{1+\theta_n + \frac{\theta_n}{n}}{1+\theta_n}
$$
Now, it remains to be seen as to why those particular terms both go to $1$. Do we require the convergence of $\theta_n$ to prove this?
Let me prove, for completeness, that $1 < \theta_n < 2$ holds for large enough $n$. That $\theta_n>1$ is obvious, since $P_{n}\left(1+\frac{1}{n}\right)<0$ by substitution. To obtain an upper bound on $\theta_n$, we assume that $n \geq 3$ and truncate the binomial expansion for $\left(1+\frac{\theta_n}{n}\right)^{2n}$ appropriately to get $$
\left(1+\frac{\theta_n}{n}\right)^{2n} > 1 + 2\theta_n
$$
So that $$
\left(1+\frac{\theta_n}{n}\right)^{2n}\left(\frac {\theta_n - 1}n\right) > \frac{(1 + 2\theta_n)(\theta_n-1)}{n} = \frac{2\theta_n^2-\theta_n-1}{n}
$$
Now, the following sequence of implications holds : $$
\frac{2\theta_n^2-\theta_n-1}{n} > \left(\frac{1+\theta_n}{n} + \frac{\theta_n}{n^2}\right) \\ \iff 2\theta_n^2-\theta_n-1 > 1+\theta_n+\frac{\theta_n}{n} \\ \impliedby 2\theta_n^2-\theta_n-1 > 1+2\theta_n \impliedby 2\theta_n^2 > 2+\frac{7}{3}\theta_n \impliedby \theta_n > 1.74.
$$
In other words, we have proved that if $\theta_n>1.74$ then $P_n\left(1+\frac{\theta_n}{n}\right)>0$. Hence, $\theta_n \leq 1.74<2$ for all $n \geq 3$, for convenience.
It is easy to handle one of the terms, in any case. Indeed, $$
1< \frac{1+\theta_n + \frac{\theta_n}{n}}{1+\theta_n} = 1+\frac{\theta_n}{1+\theta_n} \frac 1n< 1+\frac 1n
$$
and we use the squeeze theorem to conclude that $$
\lim_{n \to \infty} \frac{1+\theta_n + \frac{\theta_n}{n}}{1+\theta_n} = 1
$$ The other term is a little more subtle, and requires a suitable Taylor expansion and mean value theorem. The upshot of it all is the following inequality derived from here :$$
0 < e^x - \left(1+\frac{x}{n}\right)^n < \frac{x^2e^x}{2n}
$$
for any $n \geq 1$ and $x \in \mathbb R^+$. Dividing by $\left(1+\frac{x}{n}\right)^n$ yields $$
0< \frac{e^x}{\left(1+\frac{x}{n}\right)^n} - 1 < \frac{x^2e^x}{2n\left(1+\frac{x}{n}\right)^n} < \frac{x^2e^x}{2n}
$$
for all $n\geq 1$ and $x \in \mathbb R^+$. Setting $x = \theta_n$ for $n \geq 3$, $$
0< \frac{e^{\theta_n}}{\left(1+\frac{\theta_n}{n}\right)^n} - 1 < \frac{\theta_n^2e^{\theta_n}}{2n} < \frac{4e^2}{2n}
$$
since $\theta_n < 2$ for all $n \geq 3$. By the squeeze theorem, $$
\lim_{n \to \infty} \frac{e^{\theta_n}}{\left(1+\frac{\theta_n}{n}\right)^n} = 1
$$
and by squaring this limit,
$$
\lim_{n \to \infty} \frac{e^{2\theta_n}}{\left(1+\frac{\theta_n}{n}\right)^{2n}} = 1
$$
We have completed the two results which we need : the two terms whose limits are $1$. $u_n$ is the product of these terms, hence $\lim_{n \to \infty} u_n = 1$, as desired. As a consequence of the fact that the function $x \to e^{2x}\frac{x-1}{x+1}$ has a continuous inverse in $(1,2)$, one actually concludes that $\theta_n$ converges to the unique root $\theta \in (1,2)$ such that $e^{2\theta}\frac{\theta-1}{\theta+1} = 1$. The number $\theta \approx 1.199$ is found to be that limit.
