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Yes, because both statements are equivalent to the existence of an exact triangle of the form

The arrows are the same as in your diagram, except that you have to change of the sign of $W\rightarrow Y$ so that the first two arrows compose to $0$.

EDIT: The above 'answer' is shamely false. Let me just offer here a brief explanation of Jacob Lurie's correct answer. If the diagram is a homotopy colimit then we have an exact triangle

$$W\oplus W\stackrel{ \left(\begin{array}{cc} f&0\\ -g&g\\ 0&-h \end{array}\right) }\longrightarrow X\oplus Y\oplus Z\stackrel{(f',g',h')}\longrightarrow V\longrightarrow\Sigma W.$$

The homotopy limit condition gives rise to an exact triangle of the form

$$W\stackrel{ \left(\begin{array}{c} f\\ g\\ h \end{array}\right) }\longrightarrow X\oplus Y\oplus Z\stackrel{\left(\begin{array}{cc} f'&-g'&0\\ 0&g'&-h' \end{array}\right)}\longrightarrow V\oplus V\longrightarrow\Sigma W.$$

If $W=X=Y=0$ then you get an exact triangle by putting $V=Z$ and $h'=1_Z$ $$0\longrightarrow Z\stackrel{1}\longrightarrow Z\longrightarrow 0$$

But if you put $X=Y=0$, $V=Z$, and $h'=1_Z$, then $W=\Sigma^{-1}Z$ since the exact triangle is

$$\Sigma^{-1}Z\stackrel{0}\longrightarrow Z\stackrel{\binom{0}{1}}\longrightarrow Z\oplus Z\stackrel{(1,0)}\longrightarrow Z.$$

Yes, because both statements are equivalent to the existence of an exact triangle of the form

The arrows are the same as in your diagram, except that you have to change of the sign of $W\rightarrow Y$ so that the first two arrows compose to $0$.

EDIT: The above 'answer' is shamely false. Let me just offer here a brief explanation of Jacob Lurie's correct answer. If the diagram is a homotopy colimit then we have an exact triangle

$$W\oplus W\stackrel{ \left(\begin{array}{cc} f&0\\ -g&g\\ 0&-h \end{array}\right) }\longrightarrow X\oplus Y\oplus Z\stackrel{(f',g',h')}\longrightarrow V\longrightarrow\Sigma(W\oplus W).$$

The homotopy limit condition gives rise to an exact triangle of the form

$$W\stackrel{ \left(\begin{array}{c} f\\ g\\ h \end{array}\right) }\longrightarrow X\oplus Y\oplus Z\stackrel{\left(\begin{array}{cc} f'&-g'&0\\ 0&g'&-h' \end{array}\right)}\longrightarrow V\oplus V\longrightarrow\Sigma W.$$

If $W=X=Y=0$ then you get an exact triangle by putting $V=Z$ and $h'=1_Z$ $$0\longrightarrow Z\stackrel{1}\longrightarrow Z\longrightarrow 0$$

But if you put $X=Y=0$, $V=Z$, and $h'=1_Z$, then $W=\Sigma^{-1}Z$ since the exact triangle is

$$\Sigma^{-1}Z\stackrel{0}\longrightarrow Z\stackrel{\binom{0}{1}}\longrightarrow Z\oplus Z\stackrel{(1,0)}\longrightarrow Z.$$

Yes, because both statements are equivalent to the existence of an exact triangle of the form

The arrows are the same as in your diagram, except that you have to change of the sign of $W\rightarrow Y$ so that the first two arrows compose to $0$.

EDIT: The above 'answer' is shamely false. Let me just offer here a brief explanation of Jacob Lurie's correct answer. The diagram being a homotopy colimit is equivalent to the existence of an exact triangle

$$W\oplus W\stackrel{ \left(\begin{array}{cc} f&0\\ -g&g\\ 0&-h \end{array}\right) }\longrightarrow X\oplus Y\oplus Z\stackrel{(f',g',h')}\longrightarrow V\longrightarrow\Sigma W.$$

The homotopy limit condition would dually translate in the existence of an exact triangle of the form

$$W\stackrel{ \left(\begin{array}{c} f\\ g\\ h \end{array}\right) }\longrightarrow X\oplus Y\oplus Z\stackrel{\left(\begin{array}{cc} f'&-g'&0\\ 0&g'&-h' \end{array}\right)}\longrightarrow V\oplus V\longrightarrow\Sigma W.$$

If $W=X=Y=0$ then you get an exact triangle by putting $V=Z$ and $h'=1_Z$ $$0\longrightarrow Z\stackrel{1}\longrightarrow Z\longrightarrow 0$$

But if you put $X=Y=0$, $V=Z$, and $h'=1_Z$, then $W=\Sigma^{-1}Z$ since the exact triangle is

$$\Sigma^{-1}Z\stackrel{0}\longrightarrow Z\stackrel{\binom{0}{1}}\longrightarrow Z\oplus Z\stackrel{(1,0)}\longrightarrow Z.$$

Yes, because both statements are equivalent to the existence of an exact triangle of the form

The arrows are the same as in your diagram, except that you have to change of the sign of $W\rightarrow Y$ so that the first two arrows compose to $0$.

EDIT: The above 'answer' is shamely false. Let me just offer here a brief explanation of Jacob Lurie's correct answer. If the diagram is a homotopy colimit then we have an exact triangle

$$W\oplus W\stackrel{ \left(\begin{array}{cc} f&0\\ -g&g\\ 0&-h \end{array}\right) }\longrightarrow X\oplus Y\oplus Z\stackrel{(f',g',h')}\longrightarrow V\longrightarrow\Sigma W.$$

The homotopy limit condition gives rise to an exact triangle of the form

$$W\stackrel{ \left(\begin{array}{c} f\\ g\\ h \end{array}\right) }\longrightarrow X\oplus Y\oplus Z\stackrel{\left(\begin{array}{cc} f'&-g'&0\\ 0&g'&-h' \end{array}\right)}\longrightarrow V\oplus V\longrightarrow\Sigma W.$$

If $W=X=Y=0$ then you get an exact triangle by putting $V=Z$ and $h'=1_Z$ $$0\longrightarrow Z\stackrel{1}\longrightarrow Z\longrightarrow 0$$

But if you put $X=Y=0$, $V=Z$, and $h'=1_Z$, then $W=\Sigma^{-1}Z$ since the exact triangle is

$$\Sigma^{-1}Z\stackrel{0}\longrightarrow Z\stackrel{\binom{0}{1}}\longrightarrow Z\oplus Z\stackrel{(1,0)}\longrightarrow Z.$$

Yes, because both statements are equivalent to the existence of an exact triangle of the form

$$W\longrightarrow X\oplus Y\oplus Z\longrightarrow V\longrightarrow\Sigma W.$$

The arrows are the same as in your diagram, except that you have to change of the sign of $W\rightarrow Y$ so that the first two arrows compose to $0$.

Yes, because both statements are equivalent to the existence of an exact triangle of the form

The arrows are the same as in your diagram, except that you have to change of the sign of $W\rightarrow Y$ so that the first two arrows compose to $0$.

EDIT: The above 'answer' is shamely false. Let me just offer here a brief explanation of Jacob Lurie's correct answer. The diagram being a homotopy colimit is equivalent to the existence of an exact triangle

$$W\oplus W\stackrel{ \left(\begin{array}{cc} f&0\\ -g&g\\ 0&-h \end{array}\right) }\longrightarrow X\oplus Y\oplus Z\stackrel{(f',g',h')}\longrightarrow V\longrightarrow\Sigma W.$$

The homotopy limit condition would dually translate in the existence of an exact triangle of the form

$$W\stackrel{ \left(\begin{array}{c} f\\ g\\ h \end{array}\right) }\longrightarrow X\oplus Y\oplus Z\stackrel{\left(\begin{array}{cc} f'&-g'&0\\ 0&g'&-h' \end{array}\right)}\longrightarrow V\oplus V\longrightarrow\Sigma W.$$

If $W=X=Y=0$ then you get an exact triangle by putting $V=Z$ and $h'=1_Z$ $$0\longrightarrow Z\stackrel{1}\longrightarrow Z\longrightarrow 0$$

But if you put $X=Y=0$, $V=Z$, and $h'=1_Z$, then $W=\Sigma^{-1}Z$ since the exact triangle is

$$\Sigma^{-1}Z\stackrel{0}\longrightarrow Z\stackrel{\binom{0}{1}}\longrightarrow Z\oplus Z\stackrel{(1,0)}\longrightarrow Z.$$