This map is not the identity. It should have a graded kernel
$$I = \bigoplus_{m \ge 0} I_m \subset \bigoplus_{m \ge 0} S^m H^0(X, \mathcal{O}_X(1)),$$
which will tell you what subvariety of $\mathbb{P}^n$ is the image of the embedding $X \hookrightarrow \mathbb{P}^n$: we have $\mathop{im}{X} = V(I)$. Note also that it is certainly not necessarily the case that $H^0(X, \mathcal{O}_X(m)) \cong \mathbb{C}[s_1, \ldots, s_{n+1}]_m$.
However, the definition may feel tautological. A symmetric product of $m$ sections of $\mathcal{O}_X(1)$ should map to the product of those $m$ sections in $\mathcal{O}_X(1) \otimes \cdots \otimes \mathcal{O}_X(1) = \mathcal{O}_X(m)$:
$$
s_{i_1} \times s_{i_2} \times \cdots \times s_{i_m} \mapsto s_{i_1} s_{i_2} \cdots s_{i_m}.
$$
The kernel measures the extent to which a section of $\mathcal{O}_X(m)$ has different factorizations as a product of $m$ sections of $\mathcal{O}_X(1)$. If $t \in H^0(X, \mathcal{O}_X(m))$ is the image of both $s$ and $s' \in S^mH^0(X, \mathcal{O}_X(1))$, then $s - s' \in I$.
EDIT: I'm having trouble finding sources backing up what I'm saying here. This is from notes from a course I took 3 years ago, so I really wouldn't refer to it as "reputable." However, most comparable written sources I can find seem to prefer to restrict attention to curves, or consider the question in terms of divisors, or in terms of the map on points $X \to \mathbb{P}^n$. That said, I think the most controversial claim I make is that the product of sections of a line bundle makes sense as a section in a tensor power of that line bundle.
EDIT 2: Here I want to address numbered points in the comments.
One definition of the polynomial ring is the ring of graded functions on a finite dimensional vector space. Always, if $V$ is a complex vector space with basis $\{v_1, \ldots, v_n\}$, then $S^mV = \mathbb{C}[v_1, \ldots, v_n]_m$. This is exactly the assertion in the case $V = H^0(X, \mathcal{O}_X(1))$. You can verify this isomorphism via the the kernel of the map $TV \to \mathbb{C}[v_1, \ldots, v_n]$, if you like. The Wikipedia page on symmetric algebras has some more details, but this should make intuitive sense: The basis of degree-$m$ symmetric vectors vectors on $V$ are those of the form $v_{i_1} \times \cdots \times v_{i_m}$ where $1 \le i_1 \le \cdots \le i_m \le n$, and the basis of the degree-$m$ homogenous polynomials in $n$ variables are $x_{i_1} \cdots x_{i_m}$ where $1 \le i_1 \le \cdots \le i_m \le n$.
That's the idea, but I realize I need to give a little more detail to explain how to "multiply" global sections of a line bundle, when the global sections themselves only have the structure of a vector space. For line bundle $\mathcal{L}$ on $X$, there is a multiplication map $\bigoplus^m \mathcal{L} \to \mathcal{L}^{\otimes m}$ factoring through $S^m \mathcal{L}$. On sufficiently small open $U \subset X$, it looks like
$$
\begin{array}{ccccc}
\bigoplus_{k=1}^m \mathcal{L}(U) & \to &
S^m \mathcal{L}(U) & \to &
\mathcal{L}(U) \otimes_{\mathcal{O}_X(U)} \cdots \otimes_{\mathcal{O}_X(U)} \mathcal{L}(U) \\
(s_1|_U, \ldots, s_m|_U) &
\mapsto &
s_1|_U \times \cdots \times s_m|_U &
\mapsto &
s_1|_U \otimes \cdots \otimes s_m|_U
\end{array}
$$
Line bundles are locally isomorphic to $\mathcal{O}_X$, so this multiplication makes sense on small open $U \subset X$, and takes place in the ring $\mathcal{O}_X(U)$. Furthermore, on such $U$ there is a natural isomorphism $\mathcal{L}^{\otimes m}(U) \to \mathcal{L}(U)$, since the tensor product of an algebra with itself over itself is again that same algebra. So it makes sense to write the tensor product of sections $s_1|_U \otimes \cdots \otimes s_m|_U$ simply as an algebra product $(s_1 \cdots s_m)|_U$. Then we can cover $X$ via affine opens and glue the various sections $(s_1 \cdots s_m)|_U$ together to create a global product section $s_1 \cdots s_m \in H^0(X, \mathcal{O}_X(m))$. The fact that this gluing is allowed comes from the definition of transition maps of tensor powers—if $\{\phi_{ji}: s_i|_{U_i \cap U_j} \mapsto s_j|_{U_i \cap U_j}\}$ is the set of transition maps for a cover of $\mathcal{F}$, then $\{\phi_{ji} \otimes \cdots \otimes \phi_{ji}\}$ is the set of transition maps for a cover of $\mathcal{F}^{\otimes m}$. So the multiplication of local sections makes sense with transition maps, and we don't have to worry about non-canonicity of identifying $\mathcal{L}(U)$ with $\mathcal{O}_X(U)$. Again, unfortunately, my source is notes. I don't think this is a terribly hard issue, but I see now it has a lot of details—to get a feel for details, I don't know a better suggestion than working through an exercise-heavy textbook like Hartshorne (ch 2-4) or Vakil. For specific questions, especially related to definitions, I usually turn to the Stacks Project.
The point is that $\bigoplus_{m \ge 0} H^0(X, \mathcal{O}_X(m))$ is endowed with a structure that is different from that of $\mathbb{C}[s_1, \ldots, s_m]$. Non-uniqueness of factorization may very well exist in a graded ring whose Proj is $X$. Also, I was using "factorization" a bit loosely. If any two combinations of products of global sections
$$
\sum_{i_1 \le \cdots \le i_m} c_{i_{\bullet}} s_{i_1} \cdots s_{i_m} =
\sum_{i_{\bullet}} d_{i_{\bullet}} s_{i_1} \cdots s_{i_m} \in H^0(X, \mathcal{O}_X(m)),
$$
then I would say both have the same "factorization." Then the two preimages
$$
t = \sum_{i_{\bullet}} c_{i_{\bullet}} s_{i_1} \times \cdots \times s_{i_m},
t' = \sum_{i_{\bullet}} d_{i_{\bullet}} s_{i_1} \times \cdots \times s_{i_m}
\in S^mH^0(X, \mathcal{O}_X(1))
$$
both map to the same element and $t - t' \in I_m$. Even if the quotient is a UFD, this is possible.
Sorry to be flippant about sources. What's worked for me to understand algebraic geometry, to the extent anything has, is to slowly osmose as much understanding as I can through lectures (I've taken 3 different intro courses to algebraic geometry), looking up specific questions in Stacks Project, and working laboriously through Hartshorne problems. I really don't know which source most efficiently explains how to multiply global sections of a line bundle in terms of an affine cover, but I have developed a feeling in my bones that it's possible, and this is the way to do it.
