For each vertex $v$, define the random variable $X_v$ that is 1 if $v$ has degree $0$ and $0$ otherwise.
Then, $\mathbb{E}(X_v) = \mathbb{P}(X_v = 1)$, where the probability is the fraction of random selections of $k$ edges that leave $v$ isolated, over all possible selections of $k$ edges.
$$\mathbb{P}(X_v = 1) = \frac{\binom{\frac{n(n-1)}{2}-(n-1)}{k}}{\binom{\frac{n(n-1)}{2}}{k}} = \frac{\binom{\frac{(n-1)(n-2)}{2}}{k}}{\binom{\frac{n(n-1)}{2}}{k}}$$
Then, by linearity of expectation, the number of isolated vertices is $n\mathbb{E}(X_v)$.
Note that this does make sense for small examples. For instance if $n = 2, k= 1$, we have an expectation of $2\frac{\binom{0}{1}}{\binom{1}{1}} = 0$ isolated vertices, while for $n = 2, k=0$, the expectation is $2\frac{\binom{0}{0}}{\binom{1}{0}} = 2$. If $n = 3, k=1$, we get an expectation of one isolated vertex, etc.
Here's a sketch that I think is mostly right, though I may have made unjustified approximations that may not hold. Write $X_{ij}$ for the indicator variable that edge $(i,j)$ is isolated, where we take $i<j$. Of course $X_{ii}=0$ for all $i$. From linearity of expectation, as you said
$$
\mathbb{E}[L_n]=\sum_{i<j}\mathbb{E}[X_{ij}]={n \choose 2}p_n(1-p_n)^{2n-4}.
$$
For $p_n=\frac{a\ln n}{n}$, we have for large $n$,
$$
\mathbb{E}[L_n]= {n \choose 2}p_n(1-p_n)^{2n-4}\approx \frac{an\ln n}{2}\exp(-2a\ln n)=\frac{a}{2} n^{1-2a}\ln n.
$$
Crucially, because $a<1/2$, this is growing with $n$.
Now let's compute the variance of $L_n$. To do this, we need to compute $\mathbb{E}[X_{ij}X_{kl}]$ for various values of $i,j,k,l$. Observe that if $i=j$ and $k=l$, then $X_{ij}X_{kl}=X_{ij}^2=X_{ij}$, so we get the same computation as before. If $(i,j)$ and $(k,l)$ share a common index, it is not hard to see that $X_{ij}X_{kl}=0$, as such edges share a node so neither can be isolated. The last case is that all indices are distinct. In that case
$$
\mathbb{E}[X_{ij}X_{kl}]=\Pr(\text{$(i,j)$ is isolated and $(k,l)$ is isolated}).
$$
Considering each node among $i,j,k,l$ at a time, one can check that this happens with probability $p_n^2(1-p_n)^{4n-12}$, as there are exactly two edges we need present (namely $(i,j)$ and $(k,l)$) and exactly $4n-12$ edges we need to not be present (draw a picture if this is not clear). Note that there are exactly ${n \choose 2}{n-2 \choose 2}$ such pairs of indices that are pairwise distinct. Using these calculations, we find
\begin{align}
\mathbb{E}[L_n^2]&=\mathbb{E}\bigg[\bigg(\sum_{j<k} X_{ij}\bigg)^2\bigg]\\
&=\sum_{i<j,k<l}\mathbb{E}[X_{ij}X_{kl}]\\
&=\sum_{i<j}\mathbb{E}[X_{ij}]+\sum_{\text{$i<j,k<l$ all distinct}} \mathbb{E}[X_{ij}X_{kl}]\\
&=\mathbb{E}[L_n]+{n \choose 2}{n-2\choose 2}p_n^2(1-p_n)^{4n-12}\\
&\approx \mathbb{E}[L_n]+{n \choose 2}^2p_n^2(1-p_n)^{4n-8}\\
&=\mathbb{E}[L_n]+\mathbb{E}[L_n]^2.
\end{align}
It follows that
$$
\text{Var}[L_n]=\mathbb{E}[L_n^2]-\mathbb{E}[L_n]^2\approx \mathbb{E}[L_n].
$$
Let $\sigma=\text{Var}[L_n]\approx \sqrt{\mathbb{E}[L_n]}$. Using the Chebyshev inequality, for any fixed $k$, we thus find that
$$
\Pr(L_n\leq k)\leq \Pr\bigg(\vert L_n-\mathbb{E}[L_n]\vert> \frac{\mathbb{E}[L_n]-k}{\sigma}\sigma\bigg)\leq \frac{\text{Var}[L_n]}{(\mathbb{E}[L_n]-k)^2}=O(1/\mathbb{E}[L_n])\to 0,
$$
giving the desired statement.
Best Answer
Write $A=\sum_{i <j} X_{ij}$. Let $$p=(1+J\epsilon) \cdot \ln (n)/(2n) \quad \text{where} \quad J=\pm 1 \,.$$ Then using the Taylor expansion $\ln(1+x)=x+O(x^2)$ for $ |x|<1/2$, we infer that for $n \to \infty$ we have $$\ln E(A)=2\ln(n)+\ln(p)-2np+o(1)$$ $$= 2\ln(n)-\ln(n)+O(\ln \ln n) -(1+J\epsilon) \cdot \ln (n) =(-J\epsilon+o(1)) \cdot \ln (n) \,.$$ Thus for $J=1$, $$\mathbb{P}[G(n,p) \text{ contains an isolated edge }] \le E[A] =n^{-\epsilon+o(1)} \to 0 \quad \text{as} \quad n \to \infty \,.$$
On the other hand, if $J=-1$ then $E[A] =n^{\epsilon+o(1)}$. A direct computation shows that the covariance satisfies $$\text{Cov}(X_{ij},X_{k\ell}) \le ((1-p)^{-4}-1)E(X_{ij})E(X_{k\ell}) \le 5p n^{2\epsilon-4+o(1)} \le n^{3\epsilon-5}$$ for large $n$, so Var$(A)=o(E(A)^2).$ Chebyshev's inequality then yields that $$\mathbb{P}[G(n,p) \text{ contains no isolated edges }] \le P(A=0) \to 0 \quad \text{as} \quad n \to \infty \,.$$