For surfaces, smooth and topological classification is the same, so let me argue in the smooth category.

In Bröcker and Jänichs book "Einführung in die Differentialtopologie" (I have seen references to an English translation),
the following result is shown in full detail. Let $M_0$ and $M_1$ be two connected smooth manifolds and $D_i \subset M_i$ be two embedded discs.
The diffeomorphism type of $M_0 \sharp M_1$ (formed along the two discs) only depends on the orientation-behaviour of the discs $D_i$ (if $M_i$ is orientable)
and it does not depend on the choice of the discs if $M_i$ is nonorientable. Moreover, if both manifolds admit orientation-reversing self-diffeomorphisms,
then the diffeomorphism type of $M_0 \sharp M_1$ does not depend on any choices. As indicated by Toms comment, the question is whether any surface admits an orientation-reversing diffeomorphism.

Theorem:
''Every smooth connected orientable surface $M$ admits an orientation-reversing involution.''

Proof: the closed case is easy (by a picture), so let us assume that $M$ is open. Take a handlebody decomposition of $M$,
we can arrange it so that
$M$ has only one $0$-handle and no $2$-handles. Now $M$ is the union $D^2=M_0 \subset M_1 \subset M_2 \subset \ldots$. $M_{n+1}$ is
obtained from $M_{n}$ by gluing in a copy of $D^1 \times D^1$ along an orientation-preserving embedding $S^0 \times D^1 \to \partial M_{n}$.

Now construct the orientation-reversing diffeomorphism $f$ inductively. On $M_0 =D^2$, take a reflection. Assume that we have
constructed $f_n:M_n \to M_n$ with the extra property that $f_n$ induces the identity on $\pi_0 (\partial M_n)$ (i.e., $f_n$
does not permute the components of the boundary) and each boundary component has a parametrization by $S^1 \subset R^2$ so that
$f_{n+1}$ is given by $(x,y) \mapsto (x,-y)$ in these coordinates.
There are two cases to distinguish in the induction step.

1st case: The attaching embedding $S^0 \times D^1 \to \partial M_n$ takes the two copies of the interval into two different components. We can find an isotopic embedding $S^0 \times D^1 \to \partial M_n$ that is $Z/2$-equivariant with respect to the already
constructed $f_n$ and the self-map $(x,y) \mapsto (x,-y)$ of $D^1 \times D^1$. This is possible by the form of $f_n$ on the boundary.
The involution extends to an involution $f_{n+1}$ of $M_{n+1}$ and because the two
components where the embedding lands in are joined together, the new $f_{n+1}$ does not permute the boundary components.

2nd case: The attaching embedding takes both components of $S^0 \times D^1$ into the same component of $\partial M_{n+1}$.
This time we isotope
the attaching embedding so that it becomes equivariant with respect to $f_n$ and $(x,y) \mapsto (-x,y)$. This is possible by taking a
non-fixed point of
$f_n$ on the boundary. The component is split into two parts, but the way I have arranged the gluing guarantees that the new $f_{n+1}$ still does not
permute the components.

Note that a straightforward adaption of that argument also settles the closed case.