Note $rank(A)=rank(AX) = rank((AZ-Y)B) \le rank(B)$. Similarly $rank(B) \le rank(A)$ so $rank(A)=rank(B)$ is a necessity. To prove its sufficiency we first prove that $AZ-Y$ can be any invertible matrix. To get $AZ-Y=T$, we just need $\det(Z) \det(AZ-T) \ne 0$. We can set $Z=xI$ and note $\det(Ax-T)$ is not the zero polynomial since $\det(T)\ne 0$. Hence I need to find invertible matrices $X,T$ such that $AX=TB$ Claim 1 A matrix $K \in M_n(\mathbb{R})$ can be written as $TB$ for some invertible matrix $T$ if and only if $Row(K) = Row(B)$ Proof: Let $B = \begin{bmatrix} b_1 \\ \vdots \\ b_n \end{bmatrix}, T = \begin{bmatrix} t_{1,1} & \cdots & t_{1,n} \\ \vdots & \ddots & \vdots \\ t_{n,1} & \cdots & t_{n,n} \end{bmatrix}$, then $TB = \begin{bmatrix} t_{1,1}b_1 + \cdots + t_{1,n}b_n\\ \vdots \\ t_{n,1}b_1 + \cdots + t_{n,n}b_n \end{bmatrix}$ Hence $Row(TB) \subset Row(B)$. Similarly, $B=T^{-1}(TB)$ for some invertible $T^{-1}$, so $Row(TB) = Row(B)$. Now we prove the other direction. Suppose $Row(K)=Row(B)$ and say $\text{rank}(B)=l$. Let $b_{j_1}, \cdots, b_{j_l}$ be a basis for $Row(B)$. We can easily construct $T$ such that $TB=K$, but we need to make sure $T$ is invertible. Let $k_{i_1}, \cdots, k_{i_l}$ be a basis for $Row(K)$. We write each $k_{i_j}=\sum_t c_{j,i_t}b_{i_t}$. Note the vectors $v_j = \langle c_{j,1}, \cdots, c_{j,l}\rangle$ are linearly independent, because if not, then $k_{i_j}$'s are also linearly dependent, contradicting them being a basis. Thus we have now fixed rows $i_1,\cdots,i_l$ of $T$. Construct an arbitrary bijection $$f\colon \{1,\cdots,n\} \backslash \{i_1,\cdots, i_l\} \to \{1,\cdots,n\} \backslash \{j_1,\cdots, j_l\}$$For each row $u \in \{1,\cdots,n\} \backslash \{i_1,\cdots, i_l\}$ , let this vector have $c_{u,f(u)} = 1$, $c_{u,v}=0\forall v\in \{1,\cdots,n\} \backslash \{i_1,\cdots,i_l,f(u)\}$, and select $c_{u,i_t}$'s accordingly. Make this the $u$th row vector of $T$. It is clear that these vectors are linearly independent because row operations on these rows can be made so that row $u$ only have one nonzero entry: $c_{u,f(u)}$. Thus it suffices to construct an invertible matrix $X$ such that $Row(AX)=Row(B)$. Let rows $r_1, \cdots, r_l$ form a basis of $Row(A)$. Then $Row(AX)=span(A_p)$ where $A_p = \sum_{j=1}^n a_{r_p,j} x_j = b_{i_p}$ for $1\le p\le l$. This has a solution $(x_1,\cdots,x_n) \in (\mathbb{R}^n)^n$ where $x_1,\cdots,x_n$ are linearly independent. We are done.