By all conventions I know, this would not quite be even a fairly generalized "linear fractional transformation", for at least one technical-but-critical reason: the image is just the line $X+Y=1$. Possibly you are not asking the question you'd wish to ask.
But this map is "almost" a generalized linear fractional transformation: $g\in GL_3(\mathbb R)$ acts on $\mathbb R^3$ linearly, and then normalizing the third coordinate to $1$ gives the "projective" action of a sort mentioned by @QiaochuYuan:
$$
\pmatrix{a & b & c \cr d & e & f \cr g & h & i}\pmatrix{x \cr y & 1}
\;=\; \pmatrix{ax+by+c \cr dx+ey+f \cr gx+hy+i}\;\sim\;
\pmatrix{{ax+by+c\over gx+hy+i} \cr {dx+ey+f\over gx+hy+i} \cr 1}
$$
The problem in your example is that the associated matrix would be $\pmatrix{1 & 0 & 0 \cr 0 & 1 & 0 \cr 1 & 1 & 0}$, which is not invertible.
This leads me to wonder about your larger context, and whether you'd really want to be asking a variant question, to which the answer could be a "generalized" "yes".