**Edit.** As Friedrich Knop points out, these schemes are typically not even $S2$. I added an example at the end. (I believe it is the example that Friedrich Knop was suggesting.)

**Original answer.** This is not true. Let $d$ equal $2.$ Let $n\geq 2$ be an integer that is prime to the characteristic of $k.$ Inside $\mathbb{Z}^{\oplus 2},$ consider the subsemigroup $M$ generated by the following three elements, $$a=(1,0),\ \ b=(0,1), \ \ c = (n,-n).$$ Denote by $x$ and $y$ the elements of $k[M]$ corresponding to $a$ and $b.$ Then $k[M]$ is the $k$-subalgebra of $k[x,x^{-1},y,y^{-1}]$ generated by $x,$ by $y,$ and by $z=x^n/y^n.$ This has presentation, $$k[M] = k[x,y,z]/\langle y^nz -x^n \rangle.$$ Now consider the maximal ideal $\mathfrak{m} = \langle x,y,z-1\rangle.$ Because $n$ is prime to the characteristic, the ring extension $$k[M] \to k[M][u]/\langle (u+1)^n-z \rangle,$$ is étale near $\mathfrak{m}$. After adjoining $u$, it is clear the local ring near the maximal ideal $\mathfrak{n} = \langle x,y,z-1,u \rangle$ is not unibranch, i.e., $$k[x,y,u]/\langle (y(1+u))^n - x^n \rangle,$$ is not unibranch near $\langle x,y,u\rangle.$

I suspect that for every subsemigroup $M$ of $\mathbb{Z}^{\oplus d},$ the semigroup ring $k[M]$ is $S2.$ If the ring $k[M]$ is $S2,$ then examples such as the one above would be the only examples. Precisely, if $k[M]$ is $S2$ and if $\text{Spec}\ k[M]$ is unibranch at all codimension $1$ points, then by Hartshorne's Connectedness Theorem, the scheme is unibranch everywhere.

**Edit.** As Friedrich Knop points out, these schemes are typically not even $S2.$ For instance, let $M$ be the subsemigroup of $\mathbb{Z}_{\geq 0}^{\oplus 3}$ with the following generators, $$M = \mathbb{Z}_{\geq 0}\cdot (1,0,0) + \mathbb{Z}_{\geq 0}\cdot (0,1,0) + \mathbb{Z}_{\geq 0} \cdot (1,0,1) + \mathbb{Z}_{\geq 0}\cdot (0,1,1) + \mathbb{Z}_{\geq 0} \cdot (0,0,2).$$ Denote the respective generators of $k[M]$ as follows, $$x = \chi^{(1,0,0)}, \ y = \chi^{(0,1,0)}, \ u = \chi^{(1,0,1)}, \ v= \chi^{(0,1,1)}, \ w = \chi^{(0,0,2)}.$$ Then a presentation for the semigroup ring is, $$k[M] = k[x,y,u,v,w]/\langle u^2 - x^2w, v^2-y^2w,xv-yu,uv-xyw \rangle.$$ This is smooth if we invert either $x$, so that the fraction $z=ux^{-1}$ is in the fraction ring, or if we invert $y$, so that the fraction $z=vy^{-1}$ is in the fraction ring. Thus, the singular locus is contained in the common vanishing set of $x$ and $y$. Set-theoretically, this implies that also $u$ and $v$ equal $0$, so that the singular locus is in the affine line $\text{Spec}\ k[w].$

Since $\text{Spec}\ k[M]$ has dimension $3$ and the singular set has dimension $1$, this ring is $R1$. Yet it is not normal, since $z$ satisfies the monic polynomial $z^2 -w=0$. Therefore, by Serre's Criterion for normality, the semigroup ring $k[M]$ is not $S2.$

Please note, if you do the similar analysis as in the original example for the maximal ideal $\mathfrak{m} = \langle x,y,u,v,w-1\rangle,$ it turns out that this new example is even worse. The strict henselization is not unibranch, and it is even disconnected by removing the **codimension $2$** zero scheme of $\langle x,y,u,v\rangle.$ So it appears that we should expect no form of unibranchedness / connectedness for the strict henselizations of a non-normal semigroup ring.

1more comment