How to show that $\pi_1(\mathbb{R}^3\setminus k_{2,3}\#-k_{2,3})\not\cong\pi_1(\mathbb{R}^3\setminus k_{2,3}\# k_{2,3})$ by their presentations?
Here $k_{2,3}$ stands for the trefoil knot. $k_{2,3}\#-k_{2,3}$ stands for the connected sum of $k_{2,3}$ with its mirror image, and $k_{2,3}\# k_{2,3}$ stands for the connected sum of $k_{2,3}$ with itself.
I have no difficulty finding their presentations:
$$\pi_1(\mathbb{R}^3\setminus k_{2,3}\#-k_{2,3})\cong\langle a_1,a_2,a_3,a_4\mid a_1a_2a_1=a_2a_1a_2,\ a_3a_4a_3=a_4a_3a_4,\ a_2=a_4a_3a_4^{-1}\rangle$$ $$\pi_1(\mathbb{R}^3\setminus k_{2,3}\#k_{2,3})\cong\langle b_1,b_2,b_3,b_4\mid b_1b_2b_1=b_2b_1b_2,\ b_3b_4b_3=b_4b_3b_4,\ b_2=b_4^{-1}b_3b_4\rangle$$
But I'm in great trouble showing that they are not isomorphic. I rarely dealt with such problems, and the long complex equations are disturbing to me. By the way, they are really not isomorphic, right?
Any hint, method or solution is welcomed.
It may come as a surprise: these two groups are isomorphic!
Here's the general idea (with a concrete calculation with your presentations at the end). For a connect sum $K_1\mathbin{\#} K_2$ of knots, there is a ball $B\subset S^3$ whose boundary $\partial B$ intersects $K_1\mathbin{\#} K_2$ in exactly two points, where $B\cap (K_1\mathbin\# K_2)$ is the "$K_1$" part of the connect sum. The complement $X_1=B- \nu(K_1\mathbin\# K_2)$ is homeomorphic to $S^3-\nu(K_1)$, preserving orientations, where $\nu(K_1\mathbin\# K_2)\subset S^3$ denotes a tubular neighborhood of the connect sum. Similarly, $X_2=\overline{S^3-B}-\nu(K_1\mathbin\# K_2)$ is homeomorphic, preserving orientations, to $S^3-\nu(K_2)$. ($\overline{S^3-B}$ is homeomorphic to a closed ball.)
So, $S^3-\nu(K_1\mathbin\# K_2)=X_1\cup X_2$ with $X_1\cap X_2$ being an annulus $A$. By the van Kampen theorem, $\pi_1(S^3-\nu(K_1\mathbin\# K_2))$ is the amalgamated free product $\pi_1(X_1)*_A\pi_1(X_2)$. The image of the generator for $\pi_1(A)$ in $\pi_1(X_1)$ and $\pi_1(X_2)$ are two consistently-oriented meridians for the respective knot complement. (An "oriented meridian" is one that links positively with the knot, given the knot's orientation.)
Let $-\overline{K_2}$ indicate the mirror image of the inverse of the knot. The inverse is the knot with inverted orientation (remember: the connect sum of two knots depends on the orientation of the two knots in general --- many small knots, like the trefoil, are isotopic to their inverse). The mirror image map $S^3\to S^3$ induces an isomorphism $\pi_1(S^3-K_2)\to \pi_1(S^3-(-\overline{K_2}))$ that sends an oriented meridian to an oriented meridian.
With $\mu_1$ and $\mu_2$ being respective meridians for $K_1$ and $K_2$, $$\pi_1(S^3-\nu(K_1\mathbin\#K_2))\approx\left(\pi_1(S^3-K_1)*\pi_1(S^3-K_2)\right)/\langle \mu_1\mu_2^{-1}\rangle$$ Using the above comments about mirror images of inverses, this is also $$\left(\pi_1(S^3-K_1)*\pi_1(S^3-(-\overline{K_2}))\right)/\langle \mu_1\mu_2^{-1}\rangle\approx \pi_1(S^3-\nu(K_1\mathbin\# -\overline{K_2}))$$ In your case, the trefoil $k_{2,3}$ is isotopic to its inverse, so $$\pi_1(S^3-k_{2,3}\mathbin\#k_{2,3}) \approx \pi_1(S^3-(k_{2,3}\mathbin\#-k_{2,3})).$$ (I'm just using your notation. I think mirror images are usually denoted with an overline or by prefixing with an $m$.)
Concretely, there is an isomorphism $$\pi_1(S^3-(k_{2,3}\mathbin\#-k_{2,3})) \to \pi_1(S^3-k_{2,3}\mathbin\#k_{2,3}) $$ which we can give in the following way. Since I don't knot exactly how you got your particular presentations, I recomputed the fundamental groups:
Notice that $a_1a_3=a_3a_2=a_2a_4=a_4a_3$, so $a_1=a_4$. This corresponds to the generator of the fundamental group of the annulus separating the two knot exteriors:
Using the first three relations, we can calculate $a_1a_2a_1=a_2a_1a_2$, and with the last three, $a_1a_5a_1=a_5a_1a_5$. Hence, $$\pi_1(S^3-(k_{2,3}\mathbin\#k_{2,3}))\approx \langle a_1,a_2,a_5\mid a_1a_2a_1=a_2a_1a_2,a_1a_5a_1=a_5a_1a_5\rangle.$$ Doing the same thing for the second knot, you get $$\pi_1(S^3-(k_{2,3}\mathbin\#-k_{2,3}))\approx \langle b_1,b_2,b_5\mid b_1b_2b_1=b_2b_1b_2,b_1b_5b_1=b_5b_1b_5\rangle.$$ So the homomorphism defined by $a_1\mapsto b_1$, $a_2\mapsto b_2$, and $a_3\mapsto b_3$ is an isomorphism.
Waldhausen proved that $\pi_1(S^3-K)$, along with the peripheral subgroup generated by a meridian and a longitude, determines $K$. Your example shows why the fundamental group of the complement is not sufficient.