Are such functions differentiable?

Replace $x$ with $x+y$ to get $f(x+y)\ge f(y)(1+x)$ or $f(x+y)-f(y)\ge xf(y)$. Replace $y$ with $x+y$ and then interchange $x$ and $y$ to get $f(x+y)-f(y)\le xf(x+y)$. Together, $$ xf(y)\le f(x+y)-f(y)\le xf(x+y). $$ Dividing by $x$ and taking the limit as $x\to0$ implies that $f$ is differentiable with $f'=f$.

For any $x$ and for sufficiently large $n$ such that $1+x/n>0$, it holds that \begin{align} f(x) &\ge f\left (\frac{(n-1)x}n \right) (1+x/n)\\ &\ge f\left (\frac{(n-2)x}n \right)(1+x/n)^2 \\ &\ge \cdots \ge f(0) \left(1+ \frac x n\right)^n. \end{align} by substituting $(x,y)=(x,(n-1)x/n), ((n-1)x/n, (n-2)x/n),...$ in the given equation. In other words, $$ f(x) \ge \lim_{n\rightarrow \infty} f(0) \left(1+ \frac x n\right)^n = f(0)\cdot e^x. $$ On the other hand, for any $y$ and for sufficiently large $n$ such that $1-y/n>0$, we can similarly get the following inequality. \begin{align} f(y) &\le f\left( \frac{(n-1)y} n\right) / (1-y/n)\\ &\le \cdots \le f(0)/(1-y/n)^n. \end{align} It implies $f(y)\le f(0) \cdot e^y$. Combining these inequalities, we get that $f(x)=f(0) \cdot e^x$ is the only solution as you wanted.