Inserting $y = 0$ into the equation, we get $f(x) = f(x)g(0) + f(0)$ or
$$f(x)(1 - g(0)) = f(0).$$
If $g(0) \neq 1$ then $f(x) = \frac{f(0)}{1-g(0)}$ and $f$ is constant. This is not possible since $f$ is strictly increasing. We conclude that $g(0) = 1$ and $f(0) = 0$. From
$$f(x)g(y) + f(y) = f(x + y) = f(y)g(x) + f(x)$$
we see that $f(x)(g(y) - 1) = f(y)(g(x) - 1)$. Since $f$ is strictly increasing we have $f(x) \neq 0$ for $x \neq 0$. So, for $x, y \neq 0$ we get
$$\frac{g(y) - 1}{f(y)} = \frac{g(x) - 1}{f(x)}$$
The left-hand side of the equation only depends on $y$ and the right-hand side only depends on $x$. We conclude that both sides are constant, so $g(x) - 1 = Cf(x)$ for all $x$ other than $0$. But we have shown that $g(0) = 1$ and $f(0) = 0$, so this equality also holds for $x = 0$. Hence,
$$\begin{aligned} g(x + y) &= 1 + Cf(x + y) = 1 + Cf(x)g(y) + Cf(y) \\ &= g(y) + Cf(x)g(y) = g(y)(1 + Cf(x)) \\ &= g(y)g(x) \end{aligned}$$
for all real $x$ and $y$. This implies $g(nx) = g(x)^n$. From $g(x) = 1 - Cf(x)$ we see that $g$ is also strictly monotone (strictly increasing or strictly decreasing). Finally, $g\left(\frac{1}{x}\right) = \frac{1}{g(x)}$ and $g(0) = 1$ imply that $g(x) > 0$ for all $x \in \mathbb{R}$.
Let $g(1) = a$, where $a > 0$. Then $g(n) = a^n$ for all $n \in \mathbb{Z}$. Since $g(x) = (g(nx))^{\frac{1}{n}}$, we see that $g\left(\frac{m}{n}\right) = (g(m))^{\frac{1}{n}} = a^{\frac{m}{n}}$ or $g(x) = a^x$ for all $x \in \mathbb{Q}$. Since $g$ is strictly monotone we have $g(x) = a^x$ for all real $x$.
It is not difficult to show that the functions $g(x) = a^x$ for all $a > 0$ satisfy the conditions of the problem. If $a < 1$ let $f(x) = 1 - a^x$. If $a = 1$ let $f(x) = x$. Otherwise, let $f(x) = a^x - 1$. In all three cases $f$ is strictly increasing.