Does there exist an exponential function not vanishing at $-\infty?$
By induction, you get $f (-n )=f(-1)^{n} $. The limit condition now forces $f (-1)=1$. But then, for any $x $, $$f (-n+x)=f (-1)^nf (x)=f (x), $$ and now the limit condition gives $f (x)=1$.
The only function $f : \Bbb{R} \to \Bbb{R}$ satisfying both 1) and 2) is the constant function $f \equiv 1$.
$f$ is positive. First, if $f(a) = 0$, then $f(x) = f(x-a)f(a) = 0$ for all $x$ and thus $f(x) \to 0$ as $x \to -\infty$, which contradicts 2). So $f$ never vanishes. Then $f(x) = f(x/2)^2 > 0$ and hence $f$ is always positive.
Now let $g(x) = \log f(x)$. This functions satisfies the Cauchy functional equation $$g(x+y) = g(x) + g(y).$$ Since $g(x) \to 0$ as $x \to -\infty$, the graph of $g$ cannot be dense in $\Bbb{R}^2$ and thus $g$ is linear: $g(x) = cx$ for some constant $c$.
The only possibility for $f(x) = \mathrm{e}^{cx}$ to satisfy 2) is that $c = 0$. This corresponds to $f \equiv 1$.
(In fact, the above argument classifies all functions that satisfy 1): either $f \equiv 0$ or $\log f$ solves the Cauchy functional equation.)
$f(x)=1$ works but not at all satisfying.