Yes. Let $p=char(K)$ and $\alpha\in \overline{K(x)}\cap K((x))$ assumed to be inseparable over $K(x)$.
Let $L= K^{1/p^\infty}$ which is perfect. If $\alpha$ is inseparable over $L(x)$ then $\alpha$'s monic minimal polynomial is $f(y)=g(y^p)=h(y)^p$ with $h\in L(x)^{1/p}=L(x^{1/p})$ so $[L(x^{1/p},\alpha):L(x)] = p \deg(h) =[L(x,\alpha):L(x)]$ ie. $x^{1/p}\in L(x,\alpha)$.
This gives that $x^{1/p}\in L((x))[\alpha]= L((x))$ which is a contradiction.
Whence $\alpha$ is separable over $L(x)$.
We have $f(y)\in E[y]$ with $L/E/K$ and $E/K$ finite. Take a basis $E=\oplus_{j=1}^q b_j K$ with $b_1=1$, so that $$E(x)=\oplus_{j=1}^q b_j K(x), E((x))=\oplus_{j=1}^q b_j K((x)),\qquad f(y)=\sum_{j=1}^q b_j f_j(y)$$ with $f_j(y)\in K(x)[y]$ and $f_1(y)\ne 0$.
$\sum_{j=1}^q b_j f_j(\alpha) = 0$ in $E((x))$ gives that $$f_1(\alpha)=0$$
Whence in fact $f(y)=f_1(y)$ and the latter is separable and irreducible $\in K(x)[y]$.