Edit As Kimball point out, in the following question for me an ideal $I$ is a full $\mathbb{Z_p}$-lattice of $M_2(\mathbb{Q}_p)$ such that $$\lbrace\alpha \in M_2(\mathbb{Q}_p)\vert \alpha I \subset I \rbrace=M_2(\mathbb{Z}_P). $$
Let $p$ be a prime number and consider the ring, formed by the elements
$$ \begin{pmatrix} a & b \\ pc & d \end{pmatrix}$$
with $a,b,c,d \in \mathbb{Z}_p$. Is it true that its ideals are generated by the elements
$$ \begin{pmatrix} p^n & 0 \\ pc & p^m \end{pmatrix}$$
with $m,n$ integer and $c \in \mathbb{Z}_p$ a representative of the element $\bar c \in \mathbb{Z}/p^n\mathbb{Z}$?
How can you compute this?
It is well known that the ideals of $M_2(\mathbb{Z}_p)$ are generated by the elements
$$ \begin{pmatrix} p^n & r \\ 0 & p^m \end{pmatrix}$$
with $r \in \mathbb{Z}_p$ a representative of the element $\bar c \in \mathbb{Z}/p^m\mathbb{Z}$, and to do this it is enough look suitable $\mathbb{Z}_p$-lattices $L,M$ of a $2$-dimensional $\mathbb{Q}_p$-vector space $V$. Can I do something similar using my matrix ring?
I don't understand the "It is well known..." sentence in your question. For ideals in non-commutative rings, you need to differentiate between left and right ideals. Let $R$ be a ring and $M = M_n(R)$. Then a left ideal in $M$ is left $M$-submodule of $M$ itself. The simple submodules of $M$ are the ideals $\mathcal I_j$ for $1 \le j \le n$ consisting of the elements of $M$ which are zero off the $j$-th column. (I don't remember off the top of my head whether one need some conditions on $R$ for this, but at least for any $R$ these are ideals.)
A similar argument gives you ideals not generated by the elements you write down for your subring of $M_2(\mathbb Z_p)$.