= Solution
We first prove the stronger compact-square form, in which all the one-variable functions are continuous. There are five fixed pairs of <continuous functions> $a_q,b_q$ on the unit interval such that every real $f\in C([0,1]^2)$ has a representation
$$
\boxed{f(x,y)=\sum_{q=1}^5G_q\big(a_q(x)+b_q(y)\big),}
$$
with continuous one-variable $G_q$. We construct the inner functions and then prove convergence of the outer functions, rather than assume the <Kolmogorov-Arnold representation theorem>.
\b[Constructing separating inner functions.] Work in the <Banach space> $\mathcal X=C([0,1])^{10}$ with the maximum of the ten <supremum norms>. For each positive integer $n$, let $U_n$ consist of tuples $(a_q,b_q)_{q=1}^5$ for which there are, for each $q$, finite families of closed intervals $I_{qi}$ and $J_{qj}$ of length less than $1/n$, satisfying two conditions. Each point of the unit interval belongs to at least four of the five unions $\bigcup_iI_{qi}$, and likewise for the $J$ families. Within each fixed $q$, the compact sets
$$
K_{qij}=a_q(I_{qi})+b_q(J_{qj})
$$
are pairwise disjoint for distinct pairs $(i,j)$. Each $K_{qij}$ is a closed interval, by <continuity> and the <intermediate value theorem>. These are <grid-separated additive coordinates>.
The set $U_n$ is open: for a tuple and its finitely many witnessing intervals, the distinct compact image intervals have a positive minimum separation. A sufficiently small uniform change in the inner functions preserves that separation, while the interval-cover conditions do not change.
To prove density, start with any ten <continuous functions> and any positive approximation tolerance. Choose a common fine partition so that each original function oscillates by less than a small fraction of the tolerance on a partition cell and its immediate neighbours; choose its mesh smaller than $1/(2n)$. Near every internal division point put five small, pairwise disjoint open gaps, one for each $q$. Use these gaps to divide the interval into closed towns for family $q$, including the two endpoints in towns. Gaps from different families are disjoint, so any point misses at most one family; every town has length less than $1/n$ if the gaps are sufficiently small. Do this for both coordinates.
Approximate $a_q$ and $b_q$ by functions which are constant on their respective towns, and interpolate linearly across the gaps. The towns lie near single partition cells, so <uniform continuity> makes these approximations as close to the original functions as desired. For each $q$, perturb the finitely many town values $\alpha_{qi},\beta_{qj}$ arbitrarily slightly so that all sums $\alpha_{qi}+\beta_{qj}$ are distinct. This is possible because each unwanted equality between distinct pairs is a proper <affine hyperplane> in the <finite-dimensional vector space> of values; finitely many such hyperplanes cannot fill any <open ball>. Interpolating the perturbed values still approximates the original functions, and the rectangle images are now distinct singleton sets. Thus $U_n$ is dense.
By the <Baire category theorem>, choose one tuple in $\bigcap_{n\geq1}U_n$. To recall why completeness suffices here, start inside any <open ball> and successively choose <closed balls> of positive radii tending to zero, each contained in the preceding ball and in the next dense open $U_n$. Their centers form a <Cauchy sequence>, and its limit belongs to every ball, hence every $U_n$. Fix this tuple once and for all: it does not depend on $f$.
\b[A quantitative approximation step.] Let a continuous residual $r$ have <supremum norm> $M>0$. Choose $n$ large enough that the oscillation of $r$ on any rectangle of side lengths below $1/n$ is at most $M/20$. For each rectangle $I_{qi}\times J_{qj}$ choose a sample point $(x_{qi},y_{qj})$. Define a one-variable $h_q$ to have constant value $r(x_{qi},y_{qj})/3$ on $K_{qij}$. The image intervals are disjoint, so interpolate linearly in their gaps and extend constantly beyond the outermost ones. Then $h_q$ is continuous on the real line and
$$
\|h_q\|_\infty\leq M/3.
$$
At a given $(x,y)$, at most one family fails to contain $x$ in an $I$ town and at most one fails to contain $y$ in a $J$ town. Thus the point belongs to rectangles in $k$ families, where $3\leq k\leq5$. For each of these families the summand differs from $r(x,y)/3$ by at most $M/60$. Each remaining summand has absolute value at most $M/3$. Consequently
$$
\begin{aligned}
\left|r(x,y)-\sum_{q=1}^5h_q(a_q(x)+b_q(y))\right|
&\leq\left(\frac{k-3}{3}+\frac{5-k}{3}\right)M+\frac{kM}{60}\\
&\leq\frac23M+\frac1{12}M=\frac34M.
\end{aligned}
$$
This proves a strict contraction uniformly over the square, including points in gaps.
\b[Summing the corrections.] Set $r_0=f$ and apply that step repeatedly, subtracting the five summands at each stage. The residuals satisfy $\|r_m\|_\infty\leq(3/4)^m\|f\|_\infty$, and the corresponding corrections satisfy $\|h_{qm}\|_\infty\leq(3/4)^m\|f\|_\infty/3$. If a residual is zero, take all subsequent corrections to be zero. The series
$$
G_q=\sum_{m=0}^{\infty}h_{qm}
$$
converges uniformly on the real line to a <continuous function>. The residuals tend uniformly to zero, proving the boxed representation. An affine change of each coordinate treats any compact rectangle.
\b[Removing a restriction to a compact domain.] For a continuous real function $f$ on the whole plane, set
$$
M(t)=\max_{x^2+y^2\leq t}|f(x,y)|\quad(t\geq0).
$$
Choose a positive continuous one-variable $A$ by linear interpolation of $A(n)=(n+2)(1+M(n+1))$ at nonnegative integers. Since these values increase, for $n\leq t\leq n+1$ we have $A(t)\geq(n+2)(1+M(n+1))\geq(1+t)(1+M(t))$. Therefore
$$
b(x,y)=\frac{f(x,y)}{A(x^2+y^2)}
\quad\text{satisfies}\quad |b(x,y)|\leq\frac1{1+x^2+y^2}.
$$
Let $\sigma(x)=\tfrac12+\pi^{-1}\arctan x$. Transport $b$ to the open square by $\widetilde b(s,t)=b(\tan(\pi(s-1/2)),\tan(\pi(t-1/2)))$, and set it to zero on the square's boundary. The displayed decay estimate proves <continuity> at every boundary point, including the corners. Apply the compact-square representation to $\widetilde b$ and substitute $s=\sigma(x)$, $t=\sigma(y)$ to obtain $b$ using continuous one-variable functions and <addition>.
Finally multiply this expression by $A(x^2+y^2)$. Multiplication itself uses only <addition> and one-variable functions, because
$$
uv=\frac{(u+v)^2-(u-v)^2}{4}.
$$
Squaring, negation and scaling are one-variable operations, as are $A$ and $\sigma$. Thus the <whole-plane reduction for continuous superposition> gives a finite expression of the required kind for every continuous $f$ on the plane, without a boundedness assumption. For a complex-valued $f$, apply the argument to its real and imaginary parts and combine them using <addition> and the one-variable map $z\mapsto iz$. \b[All the constituent one-variable functions can be taken continuous.]
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