Doing this with Tietze transformations has to be the most painful way to exhibit these isomorphisms, but that's not going to stop me!
I know you said you are not interested in a geometric argument, but the below manipulations are extremely hard to find if you aren't thinking geometrically. It can be presented purely algebraically, but then it's not too different from a magic trick. I'll try to give the geometric intuition also.
I will show that $$\Delta(2,4,4) := \langle x,y,z \; \mid \; x^2 = y^2 = z^2 = 1 = (xy)^4=(yz)^4 = (xz)^2\rangle$$ is isomorphic to $$\langle a,b, r,s \; \mid \; ab = ba, s^4 = 1, s^{-1}as = b, s^{-1}bs = a^{-1}, r^2 = 1, r^{-1}ar = a, r^{-1}br = b^{-1}, (sr^2 = 1) \rangle $$ by finding Titze transformations which take one presentation to the other. I won't do $\Delta(2,3,6)$ because this is long enough already, but you should be able to use the same techniques.
First, we should notice that $\langle a, b \rangle \cong \mathbb{Z}^2$ represent the group of translations in the wallpaper group---we will think of $a$ as a translation along the $x$ axis and $b$ as a translation along the $y$-axis. Since $r^2 = 1$, we guess that $r$ is a reflection. Since it conjugates $b$ to $b^{-1}$ and commutes with $a$ we furthermore guess that it is the reflection through the $x$-axis. Likewise $s$ has order $4$, so we guess it is a rotation; it conjugates $a$ to $b$ and $b$ to $-a = a^{-1}$, so we guess it is a rotation of $\pi/2$ counterclockwise.
With these guesses in mind, we should imagine a $45$-$45$-$90$ triangle with one of its acute angles at the origin, and leg with length $1$. Then $r$ is the reflection through one of the sides, and is therefore should be $x = r$ in the presentation of $\Delta(2,4,4)$. We need to find the other sides. Let $y$ be the reflection through the hypotenuse. Then a quick geometric argument shows that $yx = s$, and hence $y = sx = sr$. Our first Tietze transformations will be introducting the generators $x, y $ along with the relations $x = r$, $y = sr$, then using those relations to eliminate $s$ and $r$ from the presentation. We now have: $$\langle a,b,x,y \; \mid \; x^2 = y^2 = 1, ab = ba, (yx)^4 = 1, xyayx = b, a^{-1} = xybyx, xax = a, xbx = b^{-1}\rangle$$
(Note that I suppressed all inverses on $x$ and $y$ since they have order 2)
Next we need to introduce the reflection, $z$, through the edge not incident to the origin. Here's a somewhat non-obvious geometric argument to see that $a = zyxy$. The translation $a$ is a composition of the reflection $z$ and the reflection through the y-axis (and the translation distance is twice the lenght of the leg). The reflection through the y-axis is just the reflection through the x-axis conjugated by a 45 degree reflection, ie $yzy$. Then $a = zyxy$.
The previous paragraph is unnecessary motivation for our Tietze transformation argument. The next step will be to introduce the new generator $z$ with the relation $z = ayxy$, or equivalently, $a = zyxy$. Using this new relation, we can eliminate $a$ from the presentation: $$\langle b,x,y,z \; \mid \; \text{see below} \rangle$$
with relations
- $x^2 = y^2 = 1$
- $zyxyb = bzyxy$
- $(yx)^4 = 1$
- $b = xyzyxyyx = xyzy$
- $yxyz^{-1} = xybyx$
- $x(zyxy)x = zyxy$
- $xbx = b^{-1}$.
Notice that we have the relation $b = xyzy$, and can therefore eliminate it from our presentation by substituting $b = xyzy$ in every relation where $b$ appears.
Relation 5 becomes $yxyz^{-1} = xyxyzyyx = xyxyzx$. Now move the $x$'s on the right to the left to get $x(yxyz^{-1})x = yxyz$. The left hand side of this is simply the inverse of the left hand side of relation 6, so we conclude $yxyz^{-1} = yxyz$, and therefore $z^{-1} = z$, ie $z^2 = 1.
Relation 6 becomes $xzyxyx = zyxy$. Shuffling around gives $xz = zyxyxyxy = z(yx)^4x = zx$, so relation 6 becomes $xz = zx$, and since $z^2 = 1$, equivalently $(xz)^2 = 1$.
Relation 2 becomes $[(zyxy)(x)]yzy = xyzyzyxy$. Cancel the right $y$'s, and use relation 6 on the bracket. Then $xzyxy yz = xyzyzyx$. Canceling some more gives $zyxz = yzyzyx$. Since $[x,z] = 1$, commute the $x$ and $z$ on the left and cancel the $x$ to give $zyz = yzyzy$, or equivalently $(zy)^4 = 1$.
Relation 7 becomes $x xyzy x = yzyx$, and canceling the $x$'s gives $yzy = yzy$, a tautology.
Then through the above Titeze transformations, we have eliminated relations 4 and 7, while replacing relations 2, 5 and 6 with $(zy)^4 = 1$, $z^2 = 1$ and $(xz)^2 = 1$ respectively. Putting it all together, the final presentation is $$\langle x,y,z \; \mid \; x^2 = y^2 = z^2 = 1 = (xz)^2 = (yx)^4 = (zy)^4 \rangle$$ as required.
To answer the question in the title: Yes, all Euclidean triangle groups are wallpaper groups. As Lee Mosher said in the comments, this statement is somewhat trial with some of the standard definitions of triangle groups/wallpaper group, so it's not entirely clear what kind of answer you want.
The triangle group $\Delta(3,3,3)$ corresponds to the wallpaper group of the hexagonal tiling.
