functional equation with two functions: $ f ( x + y ) = f ( x ) g ( y ) + f ( y ) $

If any $f(x) = 0$ or $g(y) = 0$, then your equation says $f(x+y) = f(y)$. Since $f$ is supposed to be strictly monotone, this can only happen if $x = 0$, and in particular $g$ is never $0$. Conversely, substituting $x=0$ in your equation we see that $f(0) = 0$.

Now according to your equation, if $x \ne 0$ $$ g(y) = \dfrac{f(x+y)-f(y)}{f(x)} $$ which must not depend on $x$.

Now a monotone function is almost everywhere differentiable. If $f$ is differentiable at $y$,

$$f'(y) = \lim_{h \to 0} \dfrac{f(h+y) - f(y)}{h} = g(y) \lim_{h \to 0} \dfrac{f(h) - f(0)}{h}$$

so (recalling that $g(y) \ne 0$) $f$ is differentiable at $0$, and then $f$ is differentiable everywhere, with $f'(y) = g(y) f'(0)$. Since $f'$ is not everywhere $0$, we must have $f'(0) \ne 0$ and $f'(y) \ne 0$.
Now the equation becomes $$ f(x + y) - f(y) = f(x) f'(y)/f'(0) $$

Let $F = f/f'(0)$. Then the equation becomes

$$F(x+y) - F(y) = F(x) F'(y)$$

Taking the derivative with respect to $x$,

$$F'(x+y) = F'(x) F'(y)$$

That says $F'$ is a homomorphism from the additive group $\mathbb R$ into the multiplicative group $\mathbb R_+$. The only such homomorphisms that are measurable are of the form $F'(x) = \exp(c x)$ for constant $c$. If $c = 0$ we get $F(x) = x$ and $g(y) = 1$. Otherwise, integrating, $F(x) = k + c^{-1} \exp(c x)$ for some constant $k$. Substituting this in to the original equation and rearranging, I get

$$ \left( \exp(cy)-g(y)\right) \exp(cx) = c k g(y) + \exp(cy) $$

Since the left side can't depend on $x$, we must have $g(y) = \exp(cy)$, and then $c k + 1 = 0$. Thus the "non-official" solutions are $$ \eqalign{f(x) &= \dfrac{f'(0)}{c} \left(\exp(cx) - 1\right) \cr g(y) &= \exp(cy)\cr c &\ne 0, f'(0) \ne 0}$$