Giving your series a name, I'll set $$A(P,Q) \overset{\rm def}= \exp\bigl( \log(1 + P) + \log(1 + Q) \bigr) $$
to be your power series, where $P,Q$ are free (noncommuting) variables. I'm not sure what you want to know about this series: it exists, I don't think it has a name, the first few terms are
$$ A(P,Q) = 1 + P + Q + \frac12(PQ + QP) + \frac1{3! \cdot 2}\left(-P^2 Q + 2PQP - Q^2 P-PQ^2 + 2QPQ - Q P^2\right) + $$
$$ + \frac1{4!}\left( P^3 Q - P^2QP -PQP^2 + QP^3 - PQ^2P + PQPQ + \{P\leftrightarrow Q\} \right) + \dots.$$
I don't see much of a pattern in the coefficients, and I haven't worked out the next term. If the cubic term didn't have that $\frac12$, or if the quartic terms had more fractions, I would be happier. As it should, when $[P,Q] = 0$, the series truncates to $A(P,Q) = 1 + P + Q + PQ$. It is left-right symmetric and symmetric in $P\leftrightarrow Q$.
Here's one remark that might be useful for your intended application. Let $K = k\langle\langle X,Y\rangle\rangle$ be the ring of power series in two noncommuting variables. Then there is an algebra homomorphism $\Delta: K \to K \hat\otimes K$, where $\hat\otimes$ denotes that you should complete the tensor product w.r.t. the adic topology, given on generators by $\Delta(X) = 1 \otimes X + X\otimes 1$ and $\Delta(Y) = 1 \otimes Y + Y \otimes 1$. Recall that an element $P \in K$ is primitive if $\Delta(P) = 1 \otimes P + P\otimes 1$. Then the primitive elements form a Lie subalgebra of $K$, and consist precisely of the Lie series: the power series that can be expressed without ever referring to multiplication in $K$, only to the Lie bracket $[x,y] = xy - yx$ (and that begin in degree $1$). Recall also that an element $P\in K$ is grouplike if $\Delta(P) = P\otimes P$. Then the grouplike elements if $K$ form a group under multiplication; in particular, they are all units.
If $P,Q$ are primitive, then there's no particular reason for $A(P,Q)$ to be primitive, and actually I think it never will be. However, by construction $\Delta$ is continuous w.r.t. the adic topology, and it is a homomorphism, and so $\Delta(f(P)) = f(\Delta(P))$ for any power series $f$ in one variable. Also, an element $P\in K$ is primitive iff $\exp(P)$ is grouplike. So it follows that if $1+P$ and $1+Q$ are both grouplike, then so is $A(P,Q)$.
I'll close by saying that, to me anyway, you already have an "explicit expression" for the power series, and even a "geometric interpretation", which is that it moves the additive structure of the Lie algebra to the (formal) group (whereas the BCH series moves the group multiplication to the Lie algebra). You shouldn't strongly hope for a simple description of the coefficients in terms of combinatorics, because similar descriptions for BCH, although they do exist, can be rather complicated.
Update
Above I observed that $A(P,Q)$ is not a Lie series in $P,Q$. This can be seen directly: any Lie series truncates to its linear and constant terms when $[P,Q] = 0$, whereas $A(P,Q) = (1+P)(1+Q) = 1 + P + Q + PQ$ in commuting land.
So the next best thing, as I asked in the comments, is whether
$$A(P,Q) - (1+P)(1+Q) = -\frac12[P,Q] - \frac1{12} \bigl( [P,[P,Q]] + [Q,[Q,P]] \bigr) + \dots $$
is a Lie series. Alas, this also fails, as can be seen at the quartic part. Indeed, by Jacobi and antisymmetry,
$$ [P,[Q,[P,Q]]] = [[P,Q],[P,Q]] + [Q,[P,[P,Q]] = [Q,[P,[P,Q]] = -[Q,[P,[Q,P]] $$
is the unique Lie monomial (up to scalar) of degree $P^2Q^2$. But it is antisymmetric under $P\leftrightarrow Q$, whereas $A(P,Q)$ is symmetric under the same transposition. Thus if $A(P,Q) - (1+P)(1+Q)$ were a Lie series, then it could not have any terms of degree $P^2Q^2$, whereas by direct calculation:
$$ A(P,Q) \ni \frac1{4!} \bigl( PQPQ - PQ^2P - QP^2Q + QPQP \bigr) = \frac1{24} [P,Q] ^2 $$
As I read the question, much of it amounts to, what is an abstract Lie group and what is exponentiation in a Lie group. To review, an abstract Lie group is a smooth manifold with a smooth group law, and its Lie algebra is the tangent space at the identity. The Lie bracket comes from the second derivative of the group commutator $ABA^{-1}B^{-1}$ and associativity of the group implies the Jacobi identity for its Lie algebra. For the rest I will only take finite-dimensional Lie algebras and finite-dimensional representations.
Now, you can have real or complex Lie groups and real or complex Lie algebras. Every real Lie algebra has a complexification with identical structure constants. For that reason, you may not see any distinction between a Lie algebra and its complexification in a physics treatment. But the distinction is very important, because the topology of complex Lie groups is better behaved. Note that you can also realify a Lie group or a Lie algebra, which does not change it as a set; but its real dimension is then twice its complex dimension.
There is a converse theorem that every Lie algebra integrates to an abstract Lie group. If you like, you can take it to be the universal cover, the one which is simply connected. Ado's theorem says that every Lie algebra has a faithful matrix representation. (Per the other MathOverflow question, you can make the Lie group a closed set in that faithful Lie algebra representation.) Every compact, real Lie group has a faithful, closed representation, and I suppose that every complex, simply connected Lie group has a faithful, closed, complex representation. But not every real, simply connected Lie group has a faithful representation. An important example is $\text{SL}(2,\mathbb{R})$. It has the homotopy type of a circle and therefore an infinite cyclic universal cover, but none of its cover have a faithful matrix representation. If you complexify to $\text{SL}(2,\mathbb{C})$, then it becomes simply connected.
Another important fact is that the fundamental group of a Lie group is always abelian and lifts to a subgroup of the center of its universal cover. If the center of a Lie group is a discrete subgroup, then you can divide by it to obtain a miminal Lie group model of that Lie algebra, the model that is the most rolled up.
Among the faithful representations of a real Lie algebra, I think that there is always one in which the Lie group has been unrolled as much as possible. Even without knowing that one representation, you can simply say that two points in a Lie group are equivalent if they are equal in all representations, and quotient by this equivalence. I don't know if this intermediate cover has a special name.
Okay, so that's what Lie groups are, now what about exponentiation. The most important formula for exponentiation is the "limit of compounded interest":
$$\exp(A) = \lim_{n \to \infty} \left(1+\frac{A}n\right)^{n}.$$
This formula can be approximated in an abstract Lie group (by approximating the base of the exponential to second order), so that exponentiation is well-defined in any abstract Lie group. (Or you can use the differential equation as algori says.) You can prove that this exponential satisfies the Baker-Campbell-Hausdorff formula with a positive radius of convergence. If you wanted an infinite radius of convergence, then that does not happen and there certainly are issues about radius of convergence. However, (as Theo explains) a non-zero radius of convergence is good enough to build the Lie group in patches; this is one of the ways to prove that a Lie algebra always has a Lie group.
Another formula for the exponential is the Taylor series:
$$\exp(A) = 1 + A + \frac{A^2}2 + \frac{A^3}6\cdots.$$
This formula only makes sense in a matrix representation. But, if you have a matrix representation, it agrees with the other formula.
Even in the complex group $\text{SL}(2,\mathbb{C})$, the exponential map is not surjective. You cannot reach the element
$$\begin{pmatrix} -1 & 1 \\ 0 & -1 \end{pmatrix},$$
although you can reach it and everything else in $\text{GL}(2,\mathbb{C})$. The exponential map is always dense in a complex, connected Lie group and always surjective in a compact, connected Lie group. In a real non-compact Lie group such as $\text{SL}(2,\mathbb{R})$, it isn't even dense (exercise). As Theo emphasizes, the important positive fact is that it's diffeomorphism in a neighborhood of the identity. You can walk to every point in any Lie group by taking products of elements in such a patch, and in this weaker sense every Lie group is generated by its Lie algebra.
One more remark about exponentials. Not every real Lie group has a faithful finite-dimensional representation, and it is non-trivial that every complex, simply connected Lie group does. However, a faithful infinite-dimensional continuous representation of a real Lie group is easy. If $G$ is a real Lie group, then it acts on a Hilbert space $L^2(G)$ (using its left-invariant Haar measure), which in physics-speak is the vector space of normalizable wave functions on $G$. This is a unitary representation. The Taylor series for the exponential then converges with an infinite radius of convergence. However, if you unearth the actual calculation of the exponential in this big representation, it isn't so different from the geometric exponential that algori described. Because, traveling wave functions in a manifold are only more complicated than classical trajectories in the same manifold.
Note that the unitary representations of a complex Lie group $G$ are important, but what is always meant is a representation of the realification of $G$. A representation which is both complex (in the sense that the matrix entries are complex analytic) and unitary is nonsense. In the latter case, every Lie algebra element has to have imaginary spectrum, whereas in former case, if $A$ has this property then the spectrum of $iA$ is real.
Best Answer
In the Lie Theory notes on my website based on the 2008 class by Mark Haiman, section 3.3, we give the following discussion. I should mention that I think of BCH and breaking into two parts. The first is purely algebraic:
Let $U$ be a bialgebra over field $\mathbb k$, and $s$ a formal (commuting) variable. Then we can form the $s$-adic completion $U[\![ s ]\!]$ of $U \otimes_{\mathbb k} \mathbb k[s]$. It is an algebra, of course. It is not a bialgebra in the algebraic sense. On the other hand, we can extend the comultiplication $\Delta : U \to U\otimes U$ by linearity (and continuity – it is $s$-adic-continuous) to $U[\![ s ]\!] \to (U\otimes U)[\![ s ]\!]$, and the map $U[\![ s ]\!] \otimes_{\mathbb k[\![ s ]\!]} U[\![ s ]\!] \to (U\otimes U)[\![ s ]\!]$ realizes the latter as the $s$-adic-completion of the former. So $U[\![ s ]\!]$ is a bialgebra for this $s$-adic-completed tensor product, which I will denote by $\hat\otimes$. Denote this completed comultiplication by $\hat\Delta$. Note that $\hat\Delta$ is continuous in the $s$-adic topology.
Choose $\psi \in U[\![ s ]\!]$ such that $\psi(s=0) = 0$. Then $\psi$ is primitive (i.e. $\hat\Delta(\psi) = \psi \otimes 1 + 1\otimes \psi$ iff $e^\psi$ is grouplike (i.e. $\hat\Delta(e^\psi) = e^\psi \hat\otimes e^\psi$). Note that $e^\psi = \sum_n \frac1{n!}\psi^n$ converges in the $s$-adic topology since $\psi(s=0) = 0$. Proof: By continuity, $\hat\Delta(e^\psi) = e^{\hat\Delta(\psi)}$. As a remark, note that the power series $\psi \in U[\![ s ]\!]$ is primitive iff it is primitive term-by-term.
Everything above continues to work if we use two commuting formal variables $s$ and $t$. Let $\mathfrak f$ denote the free Lie algebra on two generators $x$ and $y$. Then its universal enveloping algebra is the free associative algebra $T$ on the same two generators. (Proof: "free" functors are left adjoint to "forget to vector spaces" and "universal enveloping" is left adjoint to "forget from associative to Lie", and the composition of left adjoints is left adjoint to the corresponding composition of forgetful functors.) We will work in the (completed) bialgebra $T[\![s,t]\!]$. Define the formal series $b(sx,ty)$ by the formula $$ e^{b(sx,ty)} = e^{sx}e^{ty}. $$ Since $x$ and $y$ are primitive and the product of grouplikes is grouplike, it follows from 2. that $b(sx,ty)$ is primitive. Indeed, the coefficient on $s^mt^n$ in $b(sx,ty)$ is primitive.
The primitives in a universal enveloping algebra (in this case, $T$) are precisely the original Lie algebra (in this case, $\mathfrak f$). Thus the coefficient on $s^mt^n$ in $b(sx,ty)$ is a Lie polynomial (composition of brackets) in the noncommuting variables $x,y$. By degree counting, it is homogeneous of degree $m$ in $x$ and homogeneous of degree $n$ in $y$. More generally, $b(sx,ty)$ is a Lie series. If you work to define the correct topology so that you can talk about power series in noncommuting variables, then it makes sense to talk about the series $b(x,y)$.
The second part is manifold-theoretic:
Let $G$ denote a Lie group (whose underlying manifold is given analytic structure, and such that multiplication is an analytic map), and $\mathfrak g$ its Lie algebra. Identify the universal enveloping algebra $U\mathfrak g$ with the left-invariant differential operators on $G$. Consider the stalk $\mathcal C(G)_e$ of analytic functions defined in a neighborhood of the identity. Then $u\in U\mathfrak g = 0$ iff $u$ annihilates $\mathcal C(G)_e$.
Recall that we have a map $\exp: \mathfrak g \to G$ (defined, for example, by flowing along the left-invariant vector on $G$ field defined by $x\in \mathfrak g$). It is an analytic isomorphism near the identity, and so near the identity we have an inverse map $\log : G \to \mathfrak g$. By passing to smaller neighborhoods of the identity as needed, we can define $\beta(x,y) = \log(\exp x\exp y)$. It is a $\mathfrak g$-valued analytic function on a neighborhood of $(0,0)\in \mathfrak g \times \mathfrak g$.
Choose $f\in \mathcal C(G)_e$ and $x,y\in \mathfrak g$. Then $e^{sx} \in U\mathfrak g[\![s]\!]$ makes is $\mathbb k[\![s]\!]$-valued differential operator on $\mathcal C(G)_e$, and $e^{sx} f$ is the Taylor expansion in $s$ of $f(\exp sx)$. Similarly, $e^{sx}e^{ty}f$ computes the two-variable Taylor expansion of $f(\exp sx \exp ty)$. Let $\tilde \beta$ denote the Taylor expansion of $\beta(sx,ty)$. Then $$ e^{\tilde \beta(sx,ty)}f = f(\exp(\tilde\beta(sx,ty))) = f(\exp sx \exp ty) = e^{sx}e^{ty} f = e^{b(sx,ty)} f. $$ By 1., we must have $\tilde\beta(sx,ty) = b(sx,ty)$ as formal power series.
But $\tilde\beta$ is the Taylor expansion of the analytic function $\beta$. Thus in a small enough neighborhood of the origin, it converges. By definition, $\tilde \beta = b $ is the Baker–Campbell–Hausdorff series.