A question on certain elliptic PDE

1.If $U$ satisfies LAP then there exists a $V$ such that $(U,V)$ satisfies CR. In fact, $V$ is unique up to the addition of a term of the form $a + bx + cy + d xy$, where $a$, $b$, $c$, and $d$ are constants. This is an elementary consequence of the Frobenius Theorem.

2.You need to specify what you mean by 'algebra'. The space of functions $F = U + i V$ that satisfy your CR is infinite dimensional. In fact, any real solution to your LAP can be written in the form $$ U(x,y) = h_1(x + \omega y) + h_3(x + \omega^3 y) + h_5(x + \omega^5 y) + h_7(x + \omega^7 y) $$ where $\omega = \mathrm{e}^{i\pi/4}$ and where each $h_{2i+1}$ is holomorphic in one (complex) variable, satisfying $\overline{h_1(z)}=h_7(\bar z)$ and $\overline{h_3(z)}=h_5(\bar z)$. (I'll let you figure out the corresponding formula for $V$.) This is also provable by the Frobenius Theorem. Added remark about algebras: There are two infinite dimensional subspaces of the space of solutions of CR that are closed under multiplication: Just notice that the operator $C = {\partial_x}^2+i\, {\partial_y}^2$ factors as ${\partial_x}^2+i\, {\partial_y}^2 = \bigl({\partial_x}+\omega^3\, {\partial_y}\bigr)\bigl({\partial_x}-\omega^3\, {\partial_y}\bigr)$, and these two factors commute. Thus $C(F)=0$ is satisfied by any function $F = U+iV = h_1(x + \omega y)$ where $h_1$ is a holomorphic function of its argument (and these functions form a subalgebra) and also by any function $F = U+iV = h_5(x + \omega^5 y)$ where $h_5$ is a holomorphic function of its argument (and these functions also form a subalgebra). In fact, the space of solutions to $C(F)=0$ is almost the direct sum of these two subalgebras; they intersect in the space of constant functions and their sum is the whole space. Because these two factors also commute with $D={\partial_x}^2$, the two subalgebras are each invariant under this operator.

3.Generally, no. Just see what you have to do for surfaces. The point is that the symbol of LAP on a surface will be a quartic form that is the sum of two real, linearly independent fourth powers, and such a quartic form on a surface will determine a splitting of the tangent bundle of the surface into a sum of two line bundles. Thus, there cannot be such an operator on a compact surface with nonvanishing Euler characteristic, e.g., the $2$-sphere.

All linear functions satisfy your CR equations, so it seems fairly natural that they should be included in any related function algebra $A$. But there is no algebra of $C^1$ functions satisfying the CR equations and including a function apart from the algebra of constant functions. (I assume that the algebra is supposed to be closed under pointwise multiplication.)

Let us write $z=x+iy$. Suppose $F(z)=U(z)+iV(z)$ belongs to $A$. Now also $z\mapsto x$ is in $A$, so $f(z)=xF(z)$ is in it. Thus we have $$\begin{cases} U_{xx}=V_{yy}\\U_{yy}=-V_{xx} \end{cases}$$ and $$\begin{cases} 2U_x+xU_{xx}=xV_{yy}\\xU_{yy}=-2V_x-xV_{xx} \end{cases}.$$ We thus have $U_x=V_x=0$ and we can get the same result for $x$ replaced by $y$, whence $F$ is constant.

Remark: Suppose $F=U+iV$ and $f=u+iv$ solve the CR equations. If their product solves it too, we get $$\begin{cases} U_xu_x-V_xv_x=U_yv_y+V_yu_y\\-U_yu_y+V_yv_y=U_xv_x+V_xu_x \end{cases}.$$ Therefore any pair of functions in your algebra should satisfy these equations. I don't know if there is a nontrivial algebra if linear functions are excluded.