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To address the part of the question about a general method - seems like the Baker-Campbell-Hausdorff formula can give something tractable in this case. Denoting $e=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ and $zf=\left(\begin{smallmatrix}0&0\\z&0\end{smallmatrix}\right)$ (so that $\Delta=\operatorname{Exp}(e).\operatorname{Exp}(zf)$), one has $$ zh:=[e,zf]=z(e.f-f.e)=\left(\begin{smallmatrix}z&0\\0&-z\end{smallmatrix}\right) $$ so that the whole Lie algebra generated by $e$ and $zf$ comes out the vector space spanned by $e_n=(2z)^{n-1}e$, $f_n=2^{n-1}z^nf$, $h_n=2^{n-1}z^nh$, $n=1,2,...$, with brackets $$ [e_m,e_n]=[f_m,f_n]=[h_m,h_n]=0, $$ $$ [h_m,e_n]=2e_{m+n}, $$ $$ [h_m,f_n]=-2f_{m+n} $$ and $$ [e_m,f_n]=h_{m+n-1}. $$ In other words the Lie algebra is (almost all of the) $\mathfrak{sl}_2[z]$, its exponential map must be well studied by physicists.

In fact if I am not mistaken the BCH formula gives in this case \begin{multline*} \Delta=\operatorname{Exp}\left(e_1+f_1-\frac12h_1-\frac1{12}(e_2+f_2-\frac12h_2)+\frac1{120}(e_3+f_3-\frac12h_3)+...\right.\\\left....+(-1)^{n-1}\frac{(n-1)!}{4^{n-1}(2n-1)!!}(e_n+f_n-\frac12h_n)+...\right), \end{multline*} and since $$ e_n+f_n-\frac12h_n=\frac{(2z)^{n-1}}2\begin{pmatrix}-z&2\\2z&z\end{pmatrix}, $$ we get $$ M(z)=\frac{2\operatorname{arcsinh}\left(\frac{\sqrt z}2\right)}{\sqrt{z(z+4)}}\begin{pmatrix}-z&2\\2z&z\end{pmatrix}. $$ Although the expression differs from $M(z)$ in another answer, I believe it is actually equal.

To address the part of the question about a general method - seems like the Baker-Campbell-Hausdorff formula can give something tractable in this case. Denoting $e=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ and $zf=\left(\begin{smallmatrix}0&0\\z&0\end{smallmatrix}\right)$ (so that $\Delta=\operatorname{Exp}(e).\operatorname{Exp}(zf)$), one has $$ zh:=[e,zf]=z(e.f-f.e)=\left(\begin{smallmatrix}z&0\\0&-z\end{smallmatrix}\right) $$ so that the whole Lie algebra generated by $e$ and $zf$ comes out the vector space spanned by $e_n=(2z)^{n-1}e$, $f_n=2^{n-1}z^nf$, $h_n=2^{n-1}z^nh$, $n=1,2,...$, with brackets $$ [e_m,e_n]=[f_m,f_n]=[h_m,h_n]=0, $$ $$ [h_m,e_n]=2e_{m+n}, $$ $$ [h_m,f_n]=-2f_{m+n} $$ and $$ [e_m,f_n]=h_{m+n-1}. $$ In other words the Lie algebra is (almost all of the) $\mathfrak{sl}_2[z]$, its exponential map must be well studied by physicists.

In fact if I am not mistaken the BCH formula gives in this case \begin{multline*} \Delta=\operatorname{Exp}\left(e_1+f_1-\frac12h_1-\frac1{12}(e_2+f_2-\frac12h_2)+\frac1{120}(e_3+f_3-\frac12h_3)+...\right.\\\left....+(-1)^{n-1}\frac{(n-1)!}{4^{n-1}(2n-1)!!}(e_n+f_n-\frac12h_n)+...\right), \end{multline*} and since $$ e_n+f_n-\frac12h_n=\frac{(2z)^{n-1}}2\begin{pmatrix}-z&2\\2z&z\end{pmatrix}, $$ we get $$ M(z)=\frac12\sum_{n=1}^\infty(-1)^{n-1}\frac{(n-1)!}{(2 n-1)\text{!!}}\left(\frac z2\right)^{n-1}\begin{pmatrix}-z&2\\2z&z\end{pmatrix} $$ which (I think) boils down to the same $M(z)$ as in another answer.

To address the part of the question about a general method - seems like the Baker-Campbell-Hausdorff formula can give something tractable in this case. Denoting $e=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ and $zf=\left(\begin{smallmatrix}0&0\\z&0\end{smallmatrix}\right)$ (so that $\Delta=\operatorname{Exp}(e).\operatorname{Exp}(zf)$), one has $$ zh:=[e,zf]=z(e.f-f.e)=\left(\begin{smallmatrix}z&0\\0&-z\end{smallmatrix}\right) $$ so that the whole Lie algebra generated by $e$ and $zf$ comes out the vector space spanned by $e_n=(2z)^{n-1}e$, $f_n=2^{n-1}z^nf$, $h_n=2^{n-1}z^nh$, $n=1,2,...$, with brackets $$ [e_m,e_n]=[f_m,f_n]=[h_m,h_n]=0, $$ $$ [h_m,e_n]=2e_{m+n}, $$ $$ [h_m,f_n]=-2f_{m+n} $$ and $$ [e_m,f_n]=h_{m+n-1}. $$ In other words the Lie algebra is (almost all of the) $\mathfrak{sl}_2[z]$, its exponential map must be well studied by physicists.

In fact if I am not mistaken the BCH formula gives in this case \begin{multline*} \Delta=\operatorname{Exp}\left(e_1+f_1-\frac12h_1-\frac1{12}(e_2+f_2-\frac12h_2)+\frac1{120}(e_3+f_3-\frac12h_3)+...\right.\\\left....+(-1)^{n-1}\frac{(n-1)!}{4^{n-1}(2n-1)!!}(e_n+f_n-\frac12h_n)+...\right), \end{multline*} and since $$ e_n+f_n-\frac12h_n=\frac{(2z)^{n-1}}2\begin{pmatrix}-z&2\\2z&z\end{pmatrix}, $$ we get $$ M(z)=\frac{2\operatorname{arcsinh}\left(\frac{\sqrt z}2\right)}{\sqrt{z(z+4)}}\begin{pmatrix}-z&2\\2z&z\end{pmatrix}. $$ Note that this differs from $M(z)$ in another answer. I'm not sure but I think this one must be correct.

To address the part of the question about a general method - seems like the Baker-Campbell-Hausdorff formula can give something tractable in this case. Denoting $e=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ and $zf=\left(\begin{smallmatrix}0&0\\z&0\end{smallmatrix}\right)$ (so that $\Delta=\operatorname{Exp}(e).\operatorname{Exp}(zf)$), one has $$ zh:=[e,zf]=z(e.f-f.e)=\left(\begin{smallmatrix}z&0\\0&-z\end{smallmatrix}\right) $$ so that the whole Lie algebra generated by $e$ and $zf$ comes out the vector space spanned by $e_n=(2z)^{n-1}e$, $f_n=2^{n-1}z^nf$, $h_n=2^{n-1}z^nh$, $n=1,2,...$, with brackets $$ [e_m,e_n]=[f_m,f_n]=[h_m,h_n]=0, $$ $$ [h_m,e_n]=2e_{m+n}, $$ $$ [h_m,f_n]=-2f_{m+n} $$ and $$ [e_m,f_n]=h_{m+n-1}. $$ In other words the Lie algebra is (almost all of the) $\mathfrak{sl}_2[z]$, its exponential map must be well studied by physicists.

In fact if I am not mistaken the BCH formula gives in this case \begin{multline*} \Delta=\operatorname{Exp}\left(e_1+f_1-\frac12h_1-\frac1{12}(e_2+f_2-\frac12h_2)+\frac1{120}(e_3+f_3-\frac12h_3)+...\right.\\\left....+(-1)^{n-1}\frac{(n-1)!}{4^{n-1}(2n-1)!!}(e_n+f_n-\frac12h_n)+...\right), \end{multline*} and since $$ e_n+f_n-\frac12h_n=\frac{(2z)^{n-1}}2\begin{pmatrix}-z&2\\2z&z\end{pmatrix}, $$ we get $$ M(z)=\frac{2\operatorname{arcsinh}\left(\frac{\sqrt z}2\right)}{\sqrt{z(z+4)}}\begin{pmatrix}-z&2\\2z&z\end{pmatrix}. $$ Although the expression differs from $M(z)$ in another answer, I believe it is actually equal.

Not really an answer, I just want to address the part of the question about a general method. Seems like the Baker-Campbell-Hausdorff formula can give something tractable in this case. Denoting $e=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ and $tf=\left(\begin{smallmatrix}0&0\\t&0\end{smallmatrix}\right)$ (so that $\Delta=\operatorname{Exp}(e).\operatorname{Exp}(tf)$), one has $$ th:=[e,tf]=t(e.f-f.e)=\left(\begin{smallmatrix}t&0\\0&-t\end{smallmatrix}\right) $$ so that the whole Lie algebra generated by $e$ and $tf$ comes out the vector space spanned by $e_n=(2t)^{n-1}e$, $f_n=2^{n-1}t^nf$, $h_n=2^{n-1}t^nh$, $n=1,2,...$, with brackets $$ [e_m,e_n]=[f_m,f_n]=[h_m,h_n]=0, $$ $$ [h_m,e_n]=2e_{m+n}, $$ $$ [h_m,f_n]=-2f_{m+n} $$ and $$ [e_m,f_n]=h_{m+n-1}. $$ In other words the Lie algebra is (almost all of the) $\mathfrak{sl}_2[t]$, its exponential map must be well studied by physicists.

To address the part of the question about a general method - seems like the Baker-Campbell-Hausdorff formula can give something tractable in this case. Denoting $e=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ and $zf=\left(\begin{smallmatrix}0&0\\z&0\end{smallmatrix}\right)$ (so that $\Delta=\operatorname{Exp}(e).\operatorname{Exp}(zf)$), one has $$ zh:=[e,zf]=z(e.f-f.e)=\left(\begin{smallmatrix}z&0\\0&-z\end{smallmatrix}\right) $$ so that the whole Lie algebra generated by $e$ and $zf$ comes out the vector space spanned by $e_n=(2z)^{n-1}e$, $f_n=2^{n-1}z^nf$, $h_n=2^{n-1}z^nh$, $n=1,2,...$, with brackets $$ [e_m,e_n]=[f_m,f_n]=[h_m,h_n]=0, $$ $$ [h_m,e_n]=2e_{m+n}, $$ $$ [h_m,f_n]=-2f_{m+n} $$ and $$ [e_m,f_n]=h_{m+n-1}. $$ In other words the Lie algebra is (almost all of the) $\mathfrak{sl}_2[z]$, its exponential map must be well studied by physicists.

In fact if I am not mistaken the BCH formula gives in this case \begin{multline*} \Delta=\operatorname{Exp}\left(e_1+f_1-\frac12h_1-\frac1{12}(e_2+f_2-\frac12h_2)+\frac1{120}(e_3+f_3-\frac12h_3)+...\right.\\\left....+(-1)^{n-1}\frac{(n-1)!}{4^{n-1}(2n-1)!!}(e_n+f_n-\frac12h_n)+...\right), \end{multline*} and since $$ e_n+f_n-\frac12h_n=\frac{(2z)^{n-1}}2\begin{pmatrix}-z&2\\2z&z\end{pmatrix}, $$ we get $$ M(z)=\frac{2\operatorname{arcsinh}\left(\frac{\sqrt z}2\right)}{\sqrt{z(z+4)}}\begin{pmatrix}-z&2\\2z&z\end{pmatrix}. $$ Note that this differs from $M(z)$ in another answer. I'm not sure but I think this one must be correct.