Proof for symmetric polynomials with two variables X, Y:
Let $f(X,Y)$ be a symmetric polynomial with degree n. By definition, $f(X,Y)=f(Y,X)$.
Now, $$f(X,Y)= \sum_{0 \le i\le n,0\le j \le n,i+j\le n} a_{ij}X^iY^j$$
for some $a_{ij}$, and $a_{ij}=a_{ji} \;\forall{0\le i \le n,0\le j\le n}.\;\;\;\;(*)$
If $\alpha,\beta \in{\mathbb R}, c$ are some arbitrary constants, and $f(X,Y), g(X,Y)$ are two symmetric polynomials, then $\alpha f(X,Y)+\beta g(X,Y)+c$ is also a symmetric polynomial. This obviously extends to cases with multiple symmetric functions, i.e.
$$\sum_{j=1}^{p} \alpha_j f_j(X,Y)+c \; \text{is symmetric} \;\forall{\alpha_j \in{\mathbb R}, f_j \;\text{are all symmetric}} \;\;\;\;(**)$$
If n is even, then let $n=2m$
$$(X+Y)^n=X^n+\binom n 1 X^{n-1}Y+ \;...\;+\binom n mX^mY^m+ \;...\;+\binom n {n-1} XY^{n-1}+Y^n$$
If n is odd, then let $n=2m+1$
$$(X+Y)^n=X^n+\binom n 1 X^{n-1}Y+ \;...\;+\binom n mX^{m+1}Y^m+\binom n {m+1}X^mY^{m+1}+ \;...\;+\binom n {n-1} XY^{n-1}+Y^n$$
It is clear that $(X+Y)^n$ is symmetric $\forall{n\in {\mathbb N}}$. Moreover, $\forall 0\le k\le n,$ the $k^{th}$ term is the same as the $(n+2-k)^{th}$ term.
By $(*)$ and $(**)$ , it suffices to prove that every symmetric polynomial $f_0(X,Y)$ with at most 2 non-constant terms can be generated by $X+Y$ and $XY$, and that all polynomials with at most 2 non-constant terms generated by $X+Y$ and $XY$ are symmetric.
Suppose $\mu X^iY^j$ is symmetric, where $i\ge 0, j\ge 0$ are not both $0$. Then $X^iY^j=X^jY^i,$ which implies $i=j$.
Now, let us consider the polynomial $\;\eta_1 X^iY^j + \eta_2 X^rY^s$, where $i+j \ne r+s, i+j \ne 0, r+s \ne 0$. If it is symmetric, then $$\;\eta_1 X^iY^j + \eta_2 X^rY^s=\;\eta_1 X^jY^i + \eta_2 X^sY^r$$
This means either $X^iY^j=X^jY^i,X^rY^s=X^sY^r$ or $\eta_1=\eta_2, X^iY^j=X^sY^r, X^jY^i=X^rY^s$. Therefore, in the first case, $i=j, r=s$, and in the second case, $i=s,j=r$. Hence, all symmetric polynomials with two non-constant terms are of the form $\eta X^iY^i+\omega X^jY^j+c \;\text{or}\; \gamma (X^iY^j+X^jY^i)+c$.
Now, $\mu X^iY^i$ and $\eta X^iY^i+\omega X^jY^j$ can clearly be generated by $XY$, so we only have to check $\gamma (X^iY^j+X^jY^i)\;,$ where $i\ne j$.
Without loss of generality, suppose that $i\gt j$, so we have $$\gamma (X^iY^j+X^jY^i)=\gamma X^jY^j(X^{i-j}+Y^{i-j})$$. Now we only have to check if $$X^{i-j}+Y^{i-j}$$ can be generated by $X+Y$ and $XY$.
Let $i-j=q$. When $q=1$, $X^q+Y^q$ can trivially be generated by $X+Y$, since it's itself $X+Y$. When $q=2, X^q+Y^q=X^2+Y^2=(X+Y)^2-2XY,$ so $X^q+Y^q$ can be generated by $X+Y$ and $XY$ as well.
Notice that
$$X^{q+1}+Y^{q+1}=(X^q+Y^q)(X+Y)-(XY^q+YX^q)=(X^q+Y^q)(X+Y)-XY(X^{q-1}+Y^{q-1})$$
Therefore, $X^{q+1}+Y^{q+1}$ can be generated by $X+Y,XY,X^{q-1}+Y^{q-1},$ and $X^q+Y^q$. Hence, by induction,$\forall{q\in {\mathbb N}}, X^q+Y^q$ can be generated by $X+Y$ and $XY$. This means
$$\gamma(X^iY^j+X^jY^i)$$ can be generated by $X+Y$ and $XY$, hence all symmetric polynomials $f(X,Y)$ can be generated by $X+Y$ and $XY$.
All polynomials of degree n generated by $X+Y$ and $XY$ are of the form
$$\sum_{0 \le i'\le n,0\le j' \le n,i'+j'\le n} \kappa_{i'j'}(X+Y)^{i'}(XY)^{j'}$$
Each term of this sum is symmetric, so the sum is symmetric as well. Hence, every polynomial generated by $X+Y$ and $XY$ is symmetric, and every symmetric polynomial can be generated by $X+Y$ and $XY$.
You are correct: The elementary symmetric polynomials $e_1,\ldots,e_n$ are elements of $K$, so they are certainly not algebraically independent over $K$, and hence they cannot form a transcendence basis of anything over $K$.
The first quote shows that $L$ is algebraic over $K=k(e_1,\ldots,e_n)$ because it is the splitting field of a certain polynomial in $K[y]$. So to show that $(e_1,\ldots,e_n)$ is a transcendence basis of $L/k$ it remains to show that $(e_1,\ldots,e_n)$ is algebraically independent over $k$.
Suppose toward a contradiction that $(e_1,\ldots,e_n)$ is algebraically dependent over $k$. Then $L/k$ has a transcendence basis of fewer than $n$ elements, and so the transcendence degree of $L/k$ is less than $n$. But of course $(x_1,\ldots,x_n)$ is also a transcendence basis of $L/k$, and so the transcendence degree of $L/k$ equals $n$; a contradiction.
Best Answer
Question: "Any hint will be sufficient, I will try to prove. Also, suggest some alternate elementary way for claim, if exists."
Answer: In the case when $G:=S_2$ is the symmetric group on $x_1,x_2$, you get a direct sum decomposition as free $k[s_1,s_2]$-modules
$$A:=k[x_1,x_2]=k[s_1,s_2]\{1\}\oplus k[s_1,s_2]\{v\}:=A^G\oplus A^{-G}$$
where $v:=\frac{1}{2}(x_1-x_2), s_1:=x_1+x_2,s_2:=x_1x_2$. There is an idempotent endomorphism
$$\phi:A\rightarrow A$$
with $\phi^2=\phi$ and where $Im(\phi)=A^G\subseteq A$ is the ring of invariants. There is another idempotent $\psi$ whose image is the submodule $A^{-G}$. By definition $A^{-G}$ is the submodule of elements $x\in A$ with $\sigma(x)=-x$.
The map
$$\psi:A\rightarrow A^{-G}$$
has the property that for any element $f(x_1,x_2)\in A$ it follows
$$\psi(f)=\tilde{f}(x_1,x_2)v,$$
where $\tilde{f}(x_1,x_2)\in A^G$.
From this it follows that $dim_{k(s_1,s_2)}(k(x_1,x_2))=2$.
If $char(k)=0$ there is an operator defined as follows: If $x\in A$ let $R(x):=\frac{1}{n!}\sum_{g\in S_n}gx.$
You get a map
$$R:A \rightarrow A$$
of left $G$-modules $(G:=S_n)$, and $Im(R)=A^G$. This gives a direct sum decomposition
$$A \cong A^G \oplus Ker(R)$$
of $A$ into $G$-modules. You must study the $A^G$-module $Ker(R)$ and construct a basis as $A^G$-module. What is fun is that you can write down elementary nontrivial examples: The case of $S_3$ can be written out in complete detail. The operator $\psi$ is defined as $\psi(x):=x-R(x)$. In the case of $G:=S_2$ you may write down an explicit formula
$$\psi(x_1^ix_2^j):=\tilde{f}(x_1,x_2)v$$
where $\tilde{f}\in k[s_1,s_2]$ is an explicit $S_2$-invariant polynomial. Similarly for $S_3$.