In the book by Fulton & Harris (chapter 10, section 1) :
Finally, in the real case things are simpler: when we exponentiate the adjoint representation as above, the Lie group we arrive at is already simply connected, and so is the unique connected real Lie group with this Lie algebra.
Just to give background : It has been shown that there exist a unique non abelian two dimensional Lie algebra with basis $\{X,Y \}$ such that $[X,Y]= X.$ They calculate the adjoint form by exponenting the lie algebra to find $G_0 = \{ \begin{pmatrix} a & b\\ 0 & 1\\ \end{pmatrix} : a \ne 0 \} $. Is it not true that over real numbers $G_0$ would be homeomorphic to $\mathbb R \times S^1?$
I am not able to make sense of what exactly they are telling the above quoted paragraph.
Your formula for $G_0$ has some issues.
First, the quantity $b$ has not been specified. You seem to have assumed $b$ takes values in $S^1$, but it should instead take values in $\mathbb R$. Making that correction we get $$G_0 = \left\{ \begin{pmatrix} a & b\\ 0 & 1\\ \end{pmatrix} : a \in \mathbb R - \{0\}, \, b \in \mathbb R \right\} $$
This leads to the second issue. With that correction it follows that $G_0$ is homeomorphic to $(\mathbb R - \{0\}) \times \mathbb R$, which is not even connected, and so certainly not simply connected. But this can also be corrected, by instead specifying $a \in \mathbb R_+$: $$G_0 = \left\{ \begin{pmatrix} a & b\\ 0 & 1\\ \end{pmatrix} : a \in \mathbb R_+, \, b \in \mathbb R \right\} $$ And now $G_0$ is homeomorphic to $\mathbb R_+ \times \mathbb R$ which is indeed simply connected.
One last comment: it's true that $\mathbb R - \{0\}$ and $\mathbb R_+$ have isomorphic Lie algebras, but only the second one is simply connected. It's also true that $S^1$ and $\mathbb R$ have isomorphic Lie algebras, but only the second one is simply connected. So if you want the topological product to be simply connected, pick the simply connected factors.