The answer is probably different, depending on what you mean by the limit. (Existing of a limit as $x\to a$ when $x$ is complex is a much stronger condition then existence of a limit when $x$ is real).
Complex limits. The answer seems to be yes. What follows is a heuristic argument. Elementary functions are analytic, with at most countably many singularities. So when the finite limit exists, the singularity is removable or a ramification point. Expanding everything in power series (perhaps with fractional powers), we can compute the limit. The coefficients of these power series are elementary constants, because they are expressed in terms of derivatives. So the answer is probably yes, if we mean complex limits. (This is heuristic because singularities can accumulate so one needs more careful argument).
Real limits. The answer seems to be no. Consider a trigonometric sum $$f(t)=\sum_{j=1}^n a_j\exp(\lambda_jit),\quad i=\sqrt{-1},$$ and assume that it has no real zeros. Suppose that the $\lambda_j$ are real "elementary constants" but incommensurable. Then $f(t)=r(t)\exp(i\phi(t))$, where $r(t)>0$ and $\phi$ is a well-defined elementary real function. The limit $$m:=\lim_{t\to+\infty}\phi(t)/t$$ always exists: this is the celebrated Mean Motion Theorem. There is a formula for this limit due to A. Wintner: $$m=\sum_{j=1}^n\lambda_jW_j,\quad\mbox{where}\quad W_j=\int_{T^n}\Re\frac{a_j\exp(i\theta_j)}{\sum_ka_k\exp(i\theta_k)}d\theta_1\ldots d\theta_n,$$ where $T^n=[0,2\pi]^n$. These integrals are probably not "elementary constants": they can be expressed in Bessel functions. Again, this argument is incomplete, because one has to prove that the $W_j$ can indeed be "non-elementary" constants. (Of course you can replace $t\to\infty$ by $t\to 0$ by changing the variable to $1/t$. The difference between case 1 and case 2 is that a function can have an essential singularity and still the real limit exists).
To complete this answer one has to find specific $a_j$, evaluate the integral in terms of Bessel functions, and show that the answer is indeed not an elementary constant. This can be difficult; it is not clear what the set of elementary constants is, and how to prove that some number is not an "elementary constant". These elementary constants can be quite complicated, like $e^{\sqrt{2}\pi}$ etc.
But at least the argument shows that to compute a limit of an elementary function it is not enough to use such things as l'Hopital rule, one may need to compute definite integrals.
References on the mean motion theorem: S. Sternberg, Celestial Mechanics, Part I, Benjamin, 1969, and on the expression of $W_j$ in Bessel functions: G. Watson, Treatease on Bessel functions, CUP, 1958.
Remarks. I used the definition of "elementary function" of Liouville and Ritt (also repeated in Wikipedia). See Ritt's papers in TAMS 25 (1923) 211-222, and TAMS 27 (1925) 68-90. This definition includes the analytic continuation through removable singularities. Of course, other definitions are possible, and the answer will depend on the definition.
EDIT
- Your new definition of "elementary functions" almost coincides with what Laczkovich and Ruza called "naive elementary functions", except that he allows all constants in the definition, and you allow only $1$. I suppose that the answer to your question in this setting is also "no", but I still have to check the argument.
Ref. M. Laczkovich and I. Ruza, Elementary and integral elementary functions, Illinois J. Math., 44, 1 (2000) 161-182.