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.