$S_N(x)$ is the function $\sum_{n=0}^N \frac{x^n}{n!}$ where $x,y > 0$
My (attempted) Solution: $$S_N(x+y) = \sum_{n=0}^N \frac{(x+y)^n}{n!} = \sum_{n=0}^N\frac{{n\choose 0}x^n +{n\choose 1}x^{n-1}y +...+{n\choose n-1}xy^{n-1}+{n\choose n}y^n }{n!}$$
$$S_N(x) + S_N(y) = \sum_{n=0}^N \frac{x^n}{n!} + \sum_{n=0}^N \frac{y^n}{n!} = \sum_{n=0}^N \frac{x^n+y^n}{n!}$$
And $x^n +...+ y^n > x^n + y^n$
And done?
A recursive proof: Remark that $S'_N(x)=S_{N-1}(x)$, suppose true for $S_{N-1}$, let $y>0$ and for $x\geq 0$ define $f_N(x)=S_N(x+y)-S_N(x)-S_N(y)+1$, $f'_N(x)=S_{N-1}(x+y)-S_{N-1}(x)> S_{N-1}(y)-1>0$ since $S_{N-1}(y)>1$, so $f_N$ strictly increases, $f_N(0)=0$.