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 $f'=f$.

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 $f'=f$.

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, $$ xf(y)\le f(x+y)-f(y)\le xf(x+y). $$ This implies that $f$ is differentiable with $f'=f$.

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 $f'=f$.