Is there a continuous function with these properties (not piecewise) In short, I'm wondering if I can find a continuous non-piecewise function with these properties. I've found one that was close but not perfect. It's actually really useful to have something like this to scale it. Sorry about the lack of formatting (EDIT: Thank you)
$$\begin{aligned}
f(0) &= 1\\
f(x) &> 0\\
f'(0) &> 1\\
f'(x) &> 0\\
f''(0) &= 0\\
\lim_{x\to\infty} f '(x) &= 1 \\
\lim_{x\to -\infty} f(x) &= 0
\end{aligned}
$$
The second and fourth constraints were added afterwards, so keep that in mind while viewing comments/answers
Bonus points if you can find a function that has an elementary integral and that approaches y = x + 1
I've managed to get a function that displays all of the properties except the inflection at zero (and obviously the decrease in $f'(x)$), being hyperbolic (simply $\sqrt{1 + x^2 / 4}+ x / 2$).
The actual purpose isn't for mathematical purposes, it's for a program. In short I need to normalize a value so that it cannot drop below 0 but has no cap therein, and should approach additivity but with diminishing returns, rather than increasing that I would get with the hyperbolic one above. Thus, for most of the positive portion I need it to be concave down. If it's integrable I can do more things with it without breaking time-independence but most of the time it won't matter.

Here is an example achieving everything except for $f'(0)>1.$ It has the drawback that the parameters $a,b,c$ it seems can only be found numerically (so not an "exact" formula), but the possible advantage that perhaps by tweaking the choices of $a,b,c$ one might get $f'(0)>1.$
$$f(x)=\ln(1+a e^x)+b(\arctan(x+c)+\pi/2).$$
Note that $f,f'$ here have the required properties at $\pm \infty$, and that $f(x)>0,f'(x)>0$ provided $a,b,c$ wind up positive.
The equations $f(0)=1$ and $f''(0)=0$ can each be solved linearly for $b$, giving respectively 
$$b_1=\frac{1-\ln(a+1)}{\arctan(c)+\pi/2},\\ 
b_2=\frac{a(c^2+1)^2}{2c(a+1)^2}.$$
Since these have to be equal I experimented with the values and found that on choosing $a=.1$ that $b_1=b_2$ led to (one choice of two values) an approximate value $c \approx 0.076089...\ .$ This may then be put into either formula for $b$ to obtain $b \approx .549383...\ .$ The value of $f'(0)$ came out about $0.637$, leading me to suspect a better choice of $a,b,c$ might make $f'(0)>1$ possible to achieve.
I don't know how to include graphs, but as expected the graph of $f$ with these choices of $a,b,c$ is very flat (contact of order two with tangent line) at $x=0$. If an example could be achieved with $f'(0)>1$ it would also have this flatness, and would have to change concavity somewhere after crossing $x=0$ so its derivative would tend to $1$ as $x \to \infty.$
