Yes, $F(x)=e^x + x e^{e^x}$ satisfies an algebraic differential equation, and we can find it explicitly.

Looking at the expressions for $F$ and $F'$, we find that $$x(F'-e^x)=(1+xe^x)(F-e^x).$$

We can rewrite this equation and its derivative as \begin{align}xe^{2x}+\quad\quad\quad\ (1-x-xF)e^x &= F-xF'\\ (-1-2x)e^{2x}+(x+F+xF+xF')e^x &= xF'' \end{align}

We can solve these as linear equations for $e^x$ and $e^{2x}$, which give \begin{align} e^x&=\frac{x^2(2F'-F'')+x(F'-2F)-F}{x^2(F-F'+1)-x-1}\\ e^{2x}&=\frac{x^2(F'+FF'+F'^2-F''-FF'')+x(F''-F-F^2)-F^2}{x^2(F-F'+1)-x-1} \end{align}

Using the right hand sides in the equation $(e^x)^2=e^{2x}$ gives an algebraic differential equation for $f$.

Yes, $F(x)=e^x + x e^{e^x}$ satisfies an algebraic differential equation, and we can find it explicitly.

Looking at the expressions for $F$ and $F'$, we find that $$x(F'-e^x)=(1+xe^x)(F-e^x).$$

We can rewrite this equation and its derivative as \begin{align}xe^{2x}+\quad\quad\quad\ (1-x-xF)e^x &= F-xF'\\ (-1-2x)e^{2x}+(x+F+xF+xF')e^x &= xF'' \end{align}

We can solve these as linear equations for $e^x$ and $e^{2x}$, which give \begin{align} e^x&=\frac{x^2(2F'-F'')+x(F'-2F)-F}{x^2(F-F'+1)-x-1}\\ e^{2x}&=\frac{x^2(F'+FF'+F'^2-F''-FF'')+x(F''-F-F^2)-F^2}{x^2(F-F'+1)-x-1} \end{align}

Using the right hand sides in the equation $(e^x)^2=e^{2x}$ gives an algebraic differential equation for $F$. One way to write it is:

$$(h + 2 h x - x^2 F'')^2 =\\ \Big((1 + F) (x^2-x)-1-h x\Big) \Big(x F''(1 - x - xF)+h(2 F + h + x + xF) \Big)$$ where $h=xF'-F$.