Here's a proof, following section 4.3 of J.W. Moon's book "Counting Labelled Trees".

First, we have the following formula of Rényi (see this post for a proof and reference) for $f_k(n)$, the number of forests on $n$ vertices consisting of $k$ unrooted trees.

$$ f_k(n)= \binom nk \sum_{i=0}^k \left(-\frac12\right)^i (k+i)\,i!\, \binom{k}{i}\binom{n-k}{i} n^{n-k-i-1}.$$

Note that $U^k/k!$ is the generating function for forests of $k$ trees, so we have:

\begin{align}
\sum_{n=k}^\infty f_k(n)\frac{x^n}{n!} &= \frac{U^k}{k!}\\
&=\frac{1}{k!}(T-T^2/2)^k\\
&=\frac{1}{k!}\sum_{i=0}^k\binom{k}{i}(-1/2)^iT^{k+i}.
\end{align}

Thus $f_k(n)$ is the coefficient of $x^n$ on the right hand side. This can be determined using the following formula:

$$\frac{T^k}{k!} = \sum_{n=k}^\infty\binom{n}{k}kn^{n-k-1}\frac{x^n}{n!}.$$

To prove this, apply the Lagrange–Bürmann formula to $f(x)=T(x)$, $\phi(x)=e^T$, and $H(T)=T^k$ (see also section 4.2 of Moon's book).

To determine asymptotics, we only need to compute the coefficient of $n^{n-h}$ in the formula for $f_k(n)$. Denote this by $c(k,h)$. Then "after some simplification":

$$c(k,1)=\frac{1}{k!}\sum_{i=0}^k(-1/2)^i\binom{k}{i}(k+i)=0,$$

and

$$c(k,2)=-\frac{1}{k!}\sum_{i=0}^k(-1/2)^i(k+i)\binom{k}{i}\binom{k+i}{2}=\frac{(1/2)^{k-1}}{(k-1)!}.$$

Thus $\lim_{n\rightarrow\infty}\frac{f_k(n)}{n^{n-2}}=\frac{(1/2)^{k-1}}{(k-1)!}.$

The result follows after applying Tannery's theorem to sum these limits over all $k$:

$$\lim_{n\rightarrow\infty}\frac{N(n)}{n^{n-2}}=\sum_{k=1}^\infty\frac{(1/2)^{k-1}}{(k-1)!}=e^{1/2},$$

where $N(n)$ is the total number of rooted forests on $n$ nodes, as desired.