OK. I think EVERY (well behaved) function approaching a as limit for x towards inifinity fulfills what is asked for, because if f[ x ] -> a , then f[ x + h ] - f[ x ] -> a - a = 0 , therefore f''[ x ] -> 0 and f[ x ] + f'[ x ] ->a. f may even oscillate.

Some examples :

h = a + Sin[x]/x
Limit[h, x -> Infinity, Assumptions -> And[n > 0, x \[Element] Reals]]
Limit[D[h, x], x -> Infinity, 
 Assumptions -> And[n > 0, x \[Element] Reals]]
Limit[h + D[h, x], x -> Infinity, 
 Assumptions -> And[n > 0, x \[Element] Reals]]

h = (a x^n + x^(n - 2))/(x^n + x^(n - 3))
Limit[h, x -> Infinity, Assumptions -> And[n > 0, x \[Element] Reals]]
Limit[D[h, x], x -> Infinity, 
 Assumptions -> And[n > 0, x \[Element] Reals]]
Limit[h + D[h, x], x -> Infinity, 
 Assumptions -> And[n > 0, x \[Element] Reals]]

h = a + 1/(1 + x)^n
Limit[h, x -> Infinity, Assumptions -> And[n > 0, x \[Element] Reals]]
Limit[D[h, x], x -> Infinity, 
 Assumptions -> And[n > 0, x \[Element] Reals]]
Limit[h + D[h, x], x -> Infinity, 
 Assumptions -> And[n > 0, x \[Element] Reals]]

Here is a counterexample:

h[x_] = Exp[-x + Floor[x]];
Plot[h[x], {x, 0, 10}]
Plot[h[x] + h'[x], {x, 0, 10}]

It is essential to assume that the function is defined and differentiable on a whole half-line, in which case you can define g[x_] := x*f[Log[x]] and argue that g'[x]->c implies g[x]/x -> c too.

What about this?

fF = (f /. Flatten[DSolve[f'[x] == a - f[x], f, x]])[x]
fFp = fF + D[fF, x]
Limit[fF, x -> Infinity]
Limit[fFp, x -> Infinity]