Let $d_1= (x,y)$ and $d_2 = (u,k)$ Define $$\textrm{SwappedSpan}(d_1,d_2) = \{ ((ax + bu), (by + ak)): a,b\in\mathbb R\}$$
Now say $d_1 = (1,0)$, $d_2 = (0,1)$. Does $\textrm{SwappedSpan}(d_1,d_2)$ span $\mathbb R^2$? I think it does because using the above definition I got $\textrm{SwappedSpan}(d_1,d_2)= (a,b)$ for any real numbers $a$ and $b$. Am I missing something? It feels too easy for the problems I have been working on.
Let's understand what this function does, and what conditions must $d_{1,2}$ satisfy such that the swapped span spans $\mathbb R^2$. If you write $(a,b)$ as a column vector, you can write the swapped span as a matrix multiplication: $$\textrm{SwappedSpan}(d_1,d_2)=\begin{pmatrix}x&u\\v&y\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}$$ To span the space, it means that for any $\begin{pmatrix}\alpha\\\beta\end{pmatrix}$, the equation $$\textrm{SwappedSpan}(d_1,d_2)=\begin{pmatrix}\alpha\\\beta\end{pmatrix}$$ has a solution. That means that the first matrix in the swapped span is invertible. For that, the necessary condition is that the determinant is not $0$. With your choices, that is not the case. You either made a typo mistake in the definition when you wrote the question, or your calculation is wrong. $$(a\cdot1+b\cdot0,b\cdot 0+a\cdot1)=(a,a)\ne(a,b)$$