If you use the operators $\{+,-,\times,/\}$ (i.e., you don't included the power operator), then all of your problems are likely decidable.
Testing equality with zero
For instance, let's consider $L = \mathbb{Z} \cup \{\pi\}$. Then you can treat $\pi$ as a formal symbol, so that each leaf is a polynomial in $\mathbb{Z}[\pi]$ (e.g., the integer $5$ is the constant polynomial $5$; $\pi$ is the polynomial $\pi+0$ of degree 1). Now you can express the tree as a rational polynomial over $\mathbb{Z}$, with $\pi$ as the formal unknown.
Suppose this polynomial is $p(\pi)/q(\pi)$. Test whether $p(\pi)$ is the zero polynomial (degree $-\infty$). If it is not the zero polynomial, then the expression is not equal to zero. If $p(\pi)$ is the zero polynomial, and $q(\pi)$ is not the zero polynomial, then the expression is equal to zero. The correctness of this procedure follows from the fact that $\pi$ is transcendental.
What's the complexity of this procedure? The answer depends on the computational model. Let's assume that each operator as taking constant time to evaluate (regardless of the size of the operands). Then the complexity depends on the size of the resulting polynomials. The degree of the polynomial can grow exponentially with the depth of the tree, so if you build the polynomial recursively and express it explicitly (in coefficient form), the running time will be at most exponential in the depth of the tree. Fortunately, the degree grows at most linearly in the number of leaves in the tree, so the running time of a deterministic algorithm is linear in the size of the tree.
Therefore, assuming a straightforward representation of the tree and a simplistic computational model, this gives you a linear time algorithm for zero testing when the operators are $\{+,-,\times,/\}$.
This procedure works not just for $L=\mathbb{Z} \cup \{\pi\}$, but also for $\mathbb{N} \cup \{\pi\}$ and $\mathbb{Q} \cup \{\pi\}$.
The same procedure also works for $L=\mathbb{Q} \cup \{\pi,e\}$, if we can assume a reasonable conjecture: that $\pi$ and $e$ are algebraically independent. It is not known whether this conjecture is correct, but it seems likely. Anyway, here's the approach. We treat the polynomial as a multivariate polynomial over two unknowns $\pi,e$ instead of one unknown, but everything carries over as before, given the algebraic independence of $\pi$ and $e$. It also works for $L=\mathbb{Z} \cup \{\pi,e\}$ and $L=\mathbb{N} \cup \{\pi,e\}$, too, again, assuming the conjecture.
If you wanted to get fancy, you could use randomized algorithms for polynomial identity testing. If $L = \mathbb{Z} \cup \{\pi\}$, they'll amount to the following: choose a random prime $r$ and a random integer $s_{\pi} \in \{0,\dots,r-1\}$; replace each instance of $\pi$ with $s_{\pi}$; and then check whether the resulting expression evaluates to $0 \bmod r$. (If you have both $\pi$ and $e$, you'll pick two random integers $s_{\pi}$ and $s_e$.) You can repeat this test multiple times. If this procedure ever gives you something non-zero (modulo $r$), then the original expression is certainly non-zero. If it always gives you zero (modulo $r$), then with high probability the original expression is equal to zero. This may be more efficient in some computational models (e.g., where the time to evaluate a single operator is dependent on the size of the operands).
Sign comparison
You can also find the sign of the expression using similar procedures (again, assuming you have excluded the ^ operator, and again, assuming that $\pi$ and $e$ are algebraically independent). Evaluate the expression as a rational polynomial $p(\pi,e)/q(\pi,e)$ over $\mathbb{Q}[\pi,e]$. Assume you have determined that $p(\pi,e)/q(\pi,e) \ne 0$ and $q(\pi,e) \ne 0$. You want to know whether $p(\pi,e)/q(\pi,e) > 0$ or not.
Here is one approach. Note that $p(\pi,e)/q(\pi,e) > 0$ iff $p(\pi,e) \cdot q(\pi,e) > 0$. Therefore, we can form a new polynomial $r(\pi,e) = p(\pi,e) \cdot q(\pi,e)$ and reduce this to the problem of evaluating the sign of $r(\pi,e)$. Basically, we need to evaluate the sign of a rational polynomial in $\pi$ and $e$. We know this evaluates to something non-zero.
One approach is to compute $\pi$ and $e$ to $k$ bits of precision, and then evaluate $r(\pi,e)$ accordingly, gaining lower and upper bounds on $r(\pi,e)$. If 0 is included within this interval, double $k$, until the lower bound is strictly positive or the lower bound is strictly negative.
What's the complexity of this approach? If $|r(\pi,e)|$ evaluates to a value $\epsilon$, then I think the running time will be polynomial in the size of the input and in $\lg 1/\epsilon$.
There may be a better algorithm, but this is the best I can come up with right now.
Conclusion
The takeaway is that the power operator (^) is the real source of difficulty. Without the power operator, all of your problems can be solved without too much difficulty (assuming a reasonable conjecture).
