Question: "I want to prove that we can have a ring isomorphism: C[x,y]I≃⨁p∈C2C[x,y]Ip."
Answer: Let $A:=k[x,y]/I$ with $k$ the complex numbers. Since $dim_k(A)< \infty$ it follows $A$ is artinian and hence there is an isomorphism
$$\phi:A \cong A_1 \oplus \cdots \oplus A_d$$
where $A_i$ is an artinian local ring for every $i$ with maximal ideal $\mathfrak{m}_i$. Since $dim_k(A_i)< \infty$ it follows $\mathfrak{m}_i^{l_i}=0$ for some integer $l_i \geq 1$.
Let $\mathfrak{p}_i:=A_1\oplus \cdots \oplus \mathfrak{m}_i \oplus \cdots \oplus A_d$. It follows $A/\mathfrak{p}_i \cong A_i/\mathfrak{m}_i\cong k$ and $A/\mathfrak{p}_i^{l_i} \cong A_i$. It follows
$$\mathfrak{p}_1^{l_1}\cdots \mathfrak{p}_d^{l_d} =0.$$
By the chinese remainder theorem you get an isomorphism
$$A \cong A/(0) \cong A/\mathfrak{p}_1^{l_1}\cdots \mathfrak{p}_d^{l_d} \cong A_1 \oplus \cdots \oplus A_d$$
and since $k$ is algebraically closed it follows $\mathfrak{p}_i \cong(x-a_i,y-b_i)$ for complex numbers $(a_i,b_i)\in k^2$.
Since $\mathfrak{p}_1^{l_1}\cdots \mathfrak{p}_d^{l_d}=(0)$ in $A$ it follows
there is an equality of ideals
$$I:=(x-a_1,y-b_1)^{l_1}\cdots (x-a_d,y-b_d)^{l_d}$$
in $k[x,y]$. where $l_1,..,l_d$ is a set of integers $\geq 1$ satsifying a certain condition (see $C1$ below).
Hence if your aim is to study the Hilbert scheme, you want to parametrize the set of products of maximal ideals
$$I:=\mathfrak{p}_1^{l_1}\cdots \mathfrak{p}_d^{l_d} \subseteq k[x,y]$$
with $1 \leq d \leq n$ and $l_i \geq 1$. You will find that
$$dim_k(k[x,y]/(x,y)^i)=\binom{i+1}{2}$$
hence if $I:=(x-a_1,y-b_1)^{l_1}\cdots (x-a_d,y-b_d)^{l_d}$
it follows
$$dim_k(k[x,y]/I)= \sum_{j=1}^d \binom{l_j+1}{2}.$$
Note that $dim_k(k[x,y]/(x-a,y-b)^l)=dim_k(k[x,y]/(x,y)^l$ hence
$$dim_k(A) = \sum_j dim_k(k[x,y]/(x-a_j,y-b_j)^{l_j}=\sum_j \binom{l_j+1}{2}.$$
Hence you want to parametrize ideals $I=\mathfrak{p}_1^{l_1}\cdots \mathfrak{p}_d^{l_d}$ with $1 \leq d \leq n $ and
$$C1.\text{ }dim_k(k[x,y]/I)= \sum_{j=1}^d \binom{l_j+1}{2}=n.$$
Given an integer $1 \leq d \leq n$ you must give a combinatorial formula for the set of integers $(l_1,..,l_d)$ with $l_i\geq 1$ and $\sum_j \binom{l_j+1}{2}=n$. If you can do this it may be you are able to up give an "elementary" construction of the Hilbert scheme $Hilb^n(k[x,y])$.
Example: "I was also wondering about the ideals $I_1:=(x^4,y)$ and $I_2:=(x^3,y^2,xy)$, they are both of colength 4."
Answer: You must calculate the decomposition
$$D1.\text{ }k[x,y]/I_j \cong \oplus_i k[x,y]/(x-a_i,y-b_i)^{l_i}$$
for $j=1,2$ and check if you get the same decomposition. The rings $k[x,y]/I_j$ have finite dimension as $k$-vector spaces, and by the above argument it follows the decomposition $D1$ exist. The ideals $I_1$ and $I_2$ are equal iff they have the same decomposition. An ideal $I \subseteq k[x,y]$ have many different sets of generators and it is difficult to parametrize the set of ideals $I$ using generators. Then you must make identifications.
If you instead parametrize triples
$$\{(a_1,b_1,l_1),\ldots ,(a_d,b_d,l_d) \}\in Sym^d(k^2 \times \mathbb{Z})$$
with $(a_i,b_i,l_i)\in k^2 \times \mathbb{Z}, l_i\geq 1$ and $\sum_j \binom{l_j+1}{2}=n$
you avoid problems involved in choosing a basis for $k[x,y]/I$. Here $Sym^d(k^2 \times \mathbb{Z})$ is the "set theoretic symmetric product" of $k^2 \times \mathbb{Z}$. The symmetric group on d elements $S_d$ acts on $(k^2 \times \mathbb{Z})^d$ and $Sym^d(k^2\times \mathbb{Z}):=(k^2 \times \mathbb{Z})^d/S_d$ is the "quotient". Let
$$H^i:=Sym^i(k^2 \times \mathbb{Z})$$
and consider
$$H(n):=H^1\times H^2 \cdots \times H^n.$$
It may be you can give an "elementary" construction of the "Hilbert scheme"
$$Hilb^n(k[x,y]) \subseteq H(n).$$
Note: By "Hilbert scheme" I mean the space parametrizing ideals $I \subseteq k[x,y]$ with $dim_k(k[x,y]/I)=n$. The above "space" is not a scheme.
Note: Given a decomposition $I:=\mathfrak{m}_1^{l_1}\cdots \mathfrak{m}_d^{l_d}$ with $\mathfrak{m}_i:=(x-a_i,y-b_i)$ you get a divisor
$$D(I):= \sum_i l_i[x_i]$$
where $x_i:=Spec(k[x,y]/\mathfrak{m}_i)$. Hence this construction is related to the Hilbert-Chow morphism and divisors.
Note: In the definition of the Hilbert scheme $Hilb^n(X)$ of an affine scheme $X:=Spec(A)$ there is no Hilbert polynomial available since $X$ is affine.
