Yes, the nice simple matrices wished for in the OP's answer are, in fact, well-known 2x2 ones, leading to the answer directly without any machinery or technique.
If you wanted a name, the group is the 1-d affine group ($r\mapsto cr+d$), that of 2x2 upper triangular matrices with unit lower diagonal entry.
Observe first that the group structure is a familiar one of dilations, given the redefinitions $A\equiv X$ and $B\equiv Y -(y/x) X$, so that
$$
[A,B]=xB,
$$
so $A=z\partial_z$ and $B=z^x$, so $X=z\partial_z$ and $Y= z^x+(y/x)~z\partial_z$, so dilation maneuvers can be brought to bear.
However, the frontal, direct, approach is straightforward. Observe that
$$
A=\begin{pmatrix}
x&0\\
0&0
\end{pmatrix} , \qquad
B=\begin{pmatrix}
0&1\\
0&0
\end{pmatrix} ,
$$
satisfy the algebra, hence
$$
X=\begin{pmatrix}
x&0\\
0&0
\end{pmatrix} , \qquad
Y=\begin{pmatrix}
y&1\\
0&0
\end{pmatrix} ,
$$
so that $Y^n=y^{n-1} Y$, and consequently
$$
e^X e^Y=\begin{pmatrix}
e^x&0\\
0&1
\end{pmatrix} \begin{pmatrix}
e^y&\frac{e^y-1}{y}\\
0&1
\end{pmatrix}= \begin{pmatrix}
e^{x+y}&\frac{e^x(e^y-1)}{y}\\
0&1
\end{pmatrix}~.
$$
We just compare this to
$$
e^{aX+bY}=e^Z=\exp \left ( \begin{matrix}
e^{ax+by}&b\\
0&0
\end{matrix}\right )= \begin{pmatrix}
e^{ax+by}&\frac{b(e^{ax+by}-1)}{ax+by}\\
0&1
\end{pmatrix},
$$
the last step by virtue of $Z^n=(ax+by)^{n-1} Z$.
Hence
$$
x+y=ax+by, \qquad (ax+by) e^x(e^y-1)=by(e^{ax+by}-1),
$$
which directly determine
$$
a=\frac{(x+y)(e^x-1)}{x(e^{x+y}-1)}, \qquad b=\frac{(x+y)(e^y-1)e^x}{y(e^{x+y}-1)},
$$
as you determined.
